Is the compressibility of air the reason high aspect ratio wings are more effecient
No. In fact, compressibility does not really come into play until you come into ground effect or high Mach numbers.

Let's consider placing an object (why don't you try your (empty) coffee cup?) in mid-air. The moment you release it, it will rapidly accelerate downwards and eventually disassemble spontaneously. This is because air can move out of the way of the coffee cup. The only reason the cup experiences some resistance is due to the viscosity in air (very minor effect) and the inertia of the air that's in the way.

Lift generation relies almost solely on the inertia of the air. You accelerate air downwards, and due to the inertia of the air mass, you feel a force opposite of this, that is, upwards. 

The reason high AR wings are more efficient at this process has to do with reducing induced drag. Specifically, it is more efficient to accelerate a large mass of air a little bit, than a little bit of air a lot. Copying from an earlier answer of mine,


  The force to keep the object aloft is $$F=m_{object}g$$ The force generated by downwards momentum transfer is $$F=\dot{m}v$$ with $\dot{m}$ indicating mass flow (kilogram per second) of the air (not the mass of the object). The energy flow (power) required to  impart this momentum on the airflow is $$ P=\frac{1}{2}\dot{m}v^2$$ Here we can draw an important conclusion. The power requirement is arbitrarily small, by increasing the mass flow and decreasing the downwards velocity. 


This is the main reason for high AR wings: by increasing the wingspan (so, for equivalent surface, we increase the AR), we increase the amount of mass flow $\dot{m}$ affected by the wing.