It boils down to efficiencies of energy conversion and the cost of the technologies doing the conversions.

If you have a given mass at Earth's surface that you want in geostationary orbit, you have to raise it to the geostationary radius (or altitude, if you prefer to think in those terms), and you have to accelerate it to geostationary orbit velocity. Both of those take energy, a well-determined amount per kilogram of mass you're lofting, ~5.3 X 10$^7$ Joules per kilogram for raising to GEO radius, and ~4.7 X 10$^7$ Joules per kilogram for the orbital kinetic energy. (This is a switch from LEO. At LEO, kinetic energy is much greater than the loft energy. At GEO, loft energy is greater than kinetic energy) That is the fundamental role of both a rocket and an elevator: supply the energy to raise the mass to its desired position in Earth's gravity well, and supply the energy to get it moving with orbital velocity. The rest is "implementation details".

But as they say, "The devil is in the details."

Given the total chemical energy in a rocket's propellants, rocket engines transfer a fraction of that energy to the vehicle, and a very small fraction of that energy to the payload. The precise fraction depends on a number of factors, such as the ultimate ∆V compared to the propellants' exhaust velocity (specific impulse times Earth's gravitational acceleration), ∆V for each stage (if staged) with respect to exhaust velocity, pressure contrast between the nozzle exit and ambient—lots of things. But a typical good performance is around 10% of total available energy transferred to the entire vehicle (payload included), not to just the payload. The fraction transferred to the payload is much smaller still. The energy transferred to the vehicle itself (not the payload) is essentialy wasted. It goes to accelerating all the hardware needed to do the energy conversion (i.e., engines, tanks, pumps, feed lines, avionics, etc.) and, at any given time during the boost, all the remaining propellant. Until recently all that mass became either high-energy junk falling to Earth's surface, high-energy junk in orbit, or high-entropy gases in Earth's atmosphere. Somewhere, somebody paid—a lot!—for all that energy that's now being dissipated as heat, and all that rocket hardware that converted the chemical energy into other forms. Re-usable rocket stages are changing the balance there, but even that has costs associated, such as having to carry extra propellant to perform a landing.

Using an elevator instead, a "car" traveling up the elevator can use electricity to supply the energy to get to geostationary radius, and that energy can come from ground-based or in situ (such as solar) sources.

To get the energy needed for orbital velocity it steals energy from the elevator cable.

That doesn't come free!! More on that later.

Current ground-based electricity sources can convert chemical energy from coal, natural gas, etc. with better than 30% efficiency. None of the fuel, oxidizer (which is almost free: we get it from ambient air) or hardware to do the energy conversion need be lofted. None of the hardware involved, which for a given payload is much lighter than the rocket hardware necessary, becomes high-energy junk! So the loft part of the energy, more than half the total, comes at a much higher efficiency than is available from a rocket.

Many people assume you can get the orbital kinetic energy from the same source. Were that the case, the kinetic energy would be much more efficiently supplied by an electrical system for the elevator than by a rocket.

But the hardware accelerating the payload horizontally (perpendicular to the local vertical) is not the drive motor and drive train on the elevator car. It's the massive elevator cable itself. The higher you go on the cable, the faster it's rotating along with Earth. At the Earth's surface the payload is moving about 450 m/s. At GEO it must be moving ~3100 m/s, as the cable is. As you go up the cable, the cable's local horizontal velocity is proportional to the radius from Earth's center. So as the car moves up the cable, the cable pushes it gently in the orbital velocity direction, gradually adding horizontal kinetic energy to the car. But then the car is pushing on the cable, too, and that has consequences.

If you put a relatively small rocket engine (or some other means of applying a horizontal force) on the car you can make the net horizontal force on the cable zero. This would cancel any of the effects I'm about to discuss. But using a rocket engine you'd need to carry enough propellant for a ∆V of ~2.7 km/s, (non-trivial!) and now you're using rocket propulsion again, for nearly half of the energy needed. For now, assume no such system on the car.

The consequences: the kinetic energy the car is receiving is taken from the cable. The car slows it down—just a little at a time, but over a long time, the time it takes the car to get to GEO. This imparts a non-radial sway to the cable. The sway doesn't behave as if the cable were a rigid rod. Local displacements caused by the vertically moving car propagate as waves to the rest of the cable, vaguely akin to plucking a guitar or piano string. Eventually the cable will wind up slightly inclined westward from its anchor point, and not straight: it will be "wiggling" a little. Some of the cable's kinetic energy and some of the energy of the moving car have been converted from energy of translational motion to energy of vibration.

The non-vertical orientation is not stable. The cable, counterweight and everything connected to it will try to regain that vertical orientation. To do that, everything must accelerate eastward. Accelerating it requires energy. Where will that energy come from?

Earth's rotational energy!

If the cable is tilted slightly to the west, then the tensional force vector on the cable points mostly downward but slightly to the east. If the force on the cable has an eastward component, then the equal and opposite force on the attach point on Earth has a westward component, opposite Earth's rotation. Earth is speeding up the cable, and the cable is slowing Earth's rotation by a miniscule amount.

The cable won't descend (sag) significantly because the counterweight out there somewhere past GEO is putting enough tension on the cable that it will never fully sag, unless some idiot tries to put on a car and payload whose weight is greater than the tension force on the cable. (Aside: but as horizontal movements, i.e. differential movements with respect to the ideal radial-only position, couple into relatively small vertical forces through Coriolis forces, there will be local changes in tensile stress that allow small vertical movements. Also, small displacements from the vertical position will result in small displacements downward in Earth's gravity well; for either westward or eastward horizontal displacements, those downward movements result in small forces in an eastward direction, resulting in small corrective forces on westward displacements, small perturbing forces on eastward displacements)

The cable will be subjected to these horizontal accelerating forces all the way back to its vertical (and wiggling!) position. Which means it won't stop there. Like a simple harmonic oscillator, it continues past that equilibrium point and tilts eastward, eventually stopping and entering the reverse of the westward-tilted recovery process. In the absence of dissipatory processes (friction, flexural heating, etc.), this oscillation would continue indefinitely. There are dissipatory processes at work in a real elevator system, so eventually the cable would return to a static vertical position.

Yeah. In years, or decades, or even more, depending on the cable material. If you send cars up and down frequently and willy-nilly, not paying attention to timing or ascent/descent profiles, the vibrations and sway excited by these movements could add up to the point they overstress the cable. Needless to say, overstressing the cable is distinctly suboptimal.

How do you stop the sway in a much shorter time span?

You have to apply external forces to the cable!

Those external forces can come at least partly from the elevator car as it goes back down. If the car is carrying the same mass as it was when it went up, there will be a vertical velocity profile that will cancel the sway, and even the vibration. That doesn't mean that this theoretical velocity profile is practical. It might occasionally involve speeds greater than the car technology could handle. It might involve frequent slow-downs, even backing up, and re-accelerations, possibly making the downward trip longer than is desirable. If the optimal profile can't be implemented, then the round trip will leave the cable either swaying, or vibrating, or (most likely) both.

If the descending car's mass is different from that of the upward trip, then no doubt: there will be remnant sway and wiggle from the trip.

Properly timing and profiling another car's upward journey could also damp some of the sway and vibration.

How do you cancel remaining sway and vibration?

Again, you have to apply external forces to the cable.

As the cable approaches the equilibrium (vertical) position, you have to slow down its horizontal speed with respect to that position, so you have to apply a force in the direction opposite its motion. This works for either sway or vibration. But you have to be very careful about the combination of where and when you apply the forces. If you apply "bang-bang" forces (bang-bang control means the control force is either off or 100% on, nothing in between), say at the GEO position, you'll launch waves that travel both up and down the cable from that point, so even though you might be damping lower-frequency vibrations, you're exciting higher-frequency vibrations even more.

You could indeed apply the forces at the GEO point, but not bang-bang. They have to be applied with a profile that damps the sum of whatever motions due to vibration, traveling waves, and sway are happening at a particular time.

You could apply the forces with any system that produces translational forces in a vacuum. You could interact with Earth's magnetic fields or electric fields. That approach will take electric energy. Because you don't get to choose the direction of the magnetic field, your options for the direction of the applied force with a magnetic system are limited. Because translational force from a magnetic field requires a gradient in the field strength, and that gradient is quite small in Earth's magnetosphere, you'd need lots of electric power. Also, during magnetic storms, the magnetic field direction and strength can vary wildly, making it hard to use. There are some similar problems with the local electric field (but not the gradient-required problem), and its direction and magnitude are more variable than the magnetic field's. Either approach would take lots of electric power and somewhere, someone will have to pay for that power.

Or you could use the device most commonly used to apply translational forces in a vacuum: the rocket engine. It would have to be throttleable (no bang-bang!) or be an array of many chambers whose combined duty cycles produce a rough approximation of a continuously-variable thrust-time curve. And because Earth's gravity field is not perfectly cylindrically symmetrical, east-west oscillations would eventually couple into north-south oscillations, so you'd need engines or engine clusters pointed at the four horizontal cardinal points, not just east-west. With this approach, the cars making their rounds would have to carry with them, as part of their payload, propellants for the rocket engines. This eats into the revenue payload capacity of a car, and somewhere, someone will have to pay for the propellants and for replacing engines as they reach their operating lifetimes.

Net result: the orbital kinetic energy of the elevator car and payload does NOT come for free!

One last cost-related aspect: the cost of the technologies for generating electrical energy, distributing it to where it is needed, and converting it into kinetic energy, are significantly less than the cost of rocket technologies. Because mass is such a critical issue for rockets, much money is spent on shaving relatively small amounts of mass from components. That means the manufacture of those components is running with smaller design margins than for Earth-based systems. Running with smaller margins means more precise manufacturing methods (which are generally more expensive than less precise methods), attention to quality control with the attendant increase in inspections, documentation, etc., and the more frequent rejection of a finished part or component. All this makes a Joule of energy supplied by a rocket more expensive than one supplied by an electric motor and the generating station feeding it.

The net result is that indeed, once you have a space elevator in place (and, hoo boy, that's not a trivial task!!), the cost to orbit per kilogram for the elevator system should be considerably less than the cost for rockets. But when you consider the cable dynamics and what you have to do to control it, you find that difference probably isn't quite as large as you first thought.

The "messing with Earth's rotation" aspect makes sense when you view Earth, the elevator, the car, and the payload together as an isolated, spinning system. All that together has a certain amount of angular momentum that won't change unless acted upon by an external force. The magnitude of angular momentum is the product of rotation rate (angular velocity) and moment of inertia. (Actually angular momentum is a vector, the product of an angular velocity vector and an inertia matrix, but we don't need to get into that here!). When you run a mass from Earth's surface out to the GEO station, you increase the moment of inertia of the system by a relatively tiny fraction. Since angular momentum is constant, the angular velocity must decrease by that same tiny fraction. When the car comes back down, assuming it's carrying the same mass it did on the way up, Earth's rotation rate goes back to normal.

Hmm. Thinking about sway and vibration got me thinking about another topic, one I haven't considered before: How would lunar tides affect a space elevator? There just aren't enough hours in a day!