Why doesn't the gate voltage drop entirely across the oxide in a MOS capacitor?
The differential form of Gauss's Law is

$$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$

This means that the electric field is continuous across space. 
Sometimes when there is a highly concentrated charge (such as a surface charge on a conductor), we might obtain an approximate discontinuity in the electric field. However, if the MOSFET is designed well, no such surface charge exists at the interface between the oxide and the semiconductor, so the field cannot be discontinuous.

Therefore if the field in the oxide is non-zero, the field in the semiconductor must be nonzero as well.

Then, we also define the electrostatic potential by

$$V=-\int_C\vec{E}\cdot\mathrm{d}\vec{l}$$

So that if there is a field in the semiconductor, there must also be a variation of the potential.

Aside: In fact, it is possible for surface states to occur at this interface, which can trap electrons at the interface, resulting in a reduced field in the semiconductor. This generally leads to poorer MOSFET performance, so MOSFET makers try to avoid conditions that lead to surface states.