Could the lumpiness of the gravity of a perfectly spherical but not uniform-density airless planet be exploited to throw a stone into orbit from the surface in such a way that it stays in orbit a long time, for many revolutions, rather than crashing to the ground after less than one orbit due to returning or "trying to" return to its launch point?
100% Yes. There are orbits of the moon that take months to years to be perturbed sufficiently to impact the surface, for example PFS-1 (released by Apollo 15). Simply simulate one of these orbits and reverse it, replacing the impact with launch.
Technically, you would need to simulate the orbit with the moon spinning and orbiting in reverse, so that when flipped in time the moon would be spinning forward.
Possible if the planet itself rotates.
You first throw the stone from the densest area with above the circular orbit and a period twice the planet's own rotation.
For the next periapsis, the stone flies higher because the planet has rotated its less dense part towards it.
Gradually, the planet transfers angular momentum to the orbit of the stone.
Well, this scenario implies that the center of gravity of this planet is its geometric center, so deep density variations are expected to compensate for surface ones.
If we try to throw a stone from the surface of a perfectly spherical and uniform-density airless planet into an orbit, we need a very special elliptical orbit with a very low periapsis.
For a non uniform density we need an orbit in the equatorial plane where all mascons are compensating their influence to each other for every revolution.
An example would be a planet with three symmetric layers, constant density between 45 degrees of northern and southern lattitude and another constant density between 45 degrees of lattitude to both poles.
There might be asymmetric solutions, but these are much more difficult to find.