The key concept is that for a satellite at a fixed altitude, when the atmospheric temperatures below its altitude increase, atmospheric expansion pushes more atmosphere up above the satellite! At the satellite's altitude the pressure must increase to support the weight of that additional atmospheric mass above, and the increase in pressure outweighs the increase in temperature.
I'm going to make some simplifying assumptions that, although not descriptive of Earth's actual atmosphere, won't change the general result. I'll assume the atmosphere is isothermal, i.e. the same temperature regardless of altitude (it's not). And I'll assume that g, the acceleration due to gravity, is constant regardless of altitude (it's not). Later on I'll say why these don't change the conclusions.
I'll rearrange the Ideal Gas Law to yield mass density. The first rearrangement gives $$\frac{N}{V} =\frac{P}{RT}$$ N/V is the number of moles per volume, or the molar density. Since N is the total mass of all the molecules in the parcel, m, divided by the average molar mass $\mu$, $$\frac{m}{\mu V} =\frac{P}{RT}$$ or $$\frac{m}{V} =\frac{P\mu}{RT}$$ and m/V is just mass density.
A parameter central to atmospheric science and dynamics is the scale height, which is the vertical distance you have to travel to change the atmospheric pressure by a factor of e; e if you go downward, 1/e if you go upward. Usually denoted by H, it is given by $$H =\frac{RT}{\mu g}$$ where R is the universal gas constant, T is temperature in absolute units (like kelvins), $\mu$ is the average molar mass of the air mixture, and g is the acceleration of gravity.
Since I'm keeping T, g, and $\mu$ constant, H is a constant for this analysis.
An isothermal atmosphere has a vertical pressure profile given by $$P(h) = Po e^{-h/H}$$ where Po is the pressure at some specified altitude (like sea level), h is the altitude with respect to the specified reference altitude, H is the scale height, and P(h) is the pressure at altitude h.
Now imagine a layered, isothermal (for now) atmosphere with 10-km layers. Each layer supports all the layers above it. Assume a typical scale height for Earth's lower atmosphere of 8 km. Then at the top of a layer, the pressure would be $$P(top) = P(bottom) e^{-10/8}$$ or ~1/3.5 of the pressure at the bottom.
Now increase the temperature of the entire lowest layer by 10%, expanding it by 10% per the Ideal Gas Law, so now it's 11 km thick, with the pressure at its top unchanged. It pushed all the higher layers up by 1 km and is still supporting their weight, which hasn't changed (due to constant g).
Now do the same for the next higher layer—another 1 km rise for the layers above that one. The top of layer 2 is now 22 km up instead of the original 20 km, and everything above it has been pushed up by 2 km.
Do that another 8 times, for successively higher layers. Now the top of the 10th layer is at 110 km, where the top of the 11th layer used to be, but it's still at the original top-of-the-10th pressure. The pressure at the top of the 10th is the same as the pressure at the bottom of the 11th, so the pressure at the top of the 11th is ~1/3.5 of the pressure at the top of the 10th.
At h = 110 km, before the heating, the pressure was that of the top of level 11 (~1/3.5 times the pressure at the top of level 10), and the temperature was the original isothermal temperature, call it To. Calculating an expression for the original density at 110 km, defining Po as the original pressure at the top of level 10: $$\frac{m}{V} =\frac{(Po/3.5)\mu}{RTo} = \frac{1}{3.5} \times\frac{Po \mu}{RTo}$$
After heating, the pressure at h = 110 km is now the pressure of the top of level 10 and the temperature is up by 10%. Now calculate an expression for the density at 110 km after that heating: $$\frac{m}{V} =\frac{(Po)\mu}{R(1.1\times To)} = \frac{1}{1.1} \times\frac{Po \mu}{RTo}$$
The second is greater than the first by a factor of ~3.2! This arises from the fact that the entire mass of layer 11, which originally was beneath the 110 km altitude, is now above the 110 km level, so the pressure at 110 km must increase enough to support that added weight.
Note that this upward movement is a result of increasing the temperatures of layers below the 110 km altitude. If you increase the temperatures of layers above the specified altitude, it has no effect on the density at the specified altitude, other than a transient due to accelerating air upward.
For those who don't like the isothermal assumption, fine: make the temperature variable. Now, instead of 10 km layers, have 1 m layers, each at its own temperature, which in each layer will be very nearly constant over that 1 m altitude change. Change the temperature of each by 10% and do the expansion, and voila!: you get the same net result—after a lot more iterations through the process.
Gravitational acceleration does indeed decrease with altitude, increasing the scale height (g is in the denominator of that equation), but for an altitude change of 110 km it varies less than 4%, not enough to offset that factor-of-3+ increase seen in the analysis above.
As Mark Adler says, the reality is much more complicated, but this treatment helps to see why the density increase occurs. During real atmospheric heating events (solar flares, coronal mass ejections) the heating occurs well above the surface—nobody not in the space business ever notices it—but significant portions of it occur at altitudes below normal LEO, so it affects LEO birds.