The point of the inverted pendulum problem is that the torque on a long, thin, ground-supported object (or even just any ground-supported object with a narrow base or a hinge on its bottom) is a function of its orientation. That gives you a second order differential equation for tilt. This is NOT the case with the free-flying rocket because the engine is rotating along with the rocket. You could simulate the inverted pendulum actively, by rotating the engine so that it has a fixed orientation relative to the ground (up to the gimbaling limit, which is usually <10 degrees), but you actually want to do the opposite with the landing rocket (with significant damping to boot). Basically, you want the active control system to simulate a normal pendulum, just turned upside down (but with the damping so that it doesn't actually oscillate).
> The rocket is unstable in position and inclination (attitude).
Not the latter unless you make it behave in that way. You could also check on the Apollo LM landing guidance, maybe it will become more clear if you study that.
> So you're right that there's more to it, but it's still the inverted pendulum problem for attitude
No, it's not. That's the point.
> Are you saying Bezos is lying in this post when he explicitly says "When you do a vertical landing, you’re solving the classic inverted pendulum problem"? He would know, and I don't see a reason for him to lie about it.
No, I'm saying that this phrasing makes no sense. Remember that Goddard made a similar mistake when reasoning about rockets and pendulums, so arguments to authority are demonstrably to be avoided here. A significant clarification on his part could make it more comprehensible what he actually meant, but making any kind of analogy to a ground-supported object is clearly bad when talking about a free-flying body.