This derivation assumes the container is a cube, but the law holds for containers of any shape, right?
The resulting law holds for any shape of container. However, the details of the derivation depend on the geometry. A cube is easy, so we use it.
This derivation assumes each particle has the same mass, but the law holds for a mixture of gasses of different masses, like air, right?
Yes, it holds for a mixture of any ideal gases - the key is that only the kinetic energy of the gas molecules matters. Heavier molecules would be moving slower at the same temperature, but they would exert the same force (and therefore pressure) on the walls.
The derivation assumes that the particle intercepts a give side with frequency 2L/vx. Why can we make this assumption?
This assumption works for only one particular case. If you notice the third sentence at the beginning of the pressure derivation (emphasis is mine):
When a gas molecule collides with the wall of the container perpendicular to the x coordinate axis and bounces off in the opposite direction with the same speed
They are looking at the special case of a collision at a 90 degree angle, which bounces off the opposite wall. In that case, it will take the particle t = L/V time to reach the opposite wall, and t = L/V time to bounce back - so the total is 2L/V. This illustration may help explain it better:
Now this might sound like an unreasonably strict condition, but the reason that it works is that the component velocity distribution for each particle is symmetrical (it's a normal distribution), which means that for every particle that impacts the wall at 95 degrees (for example) there is another that impacts 85 degrees. As a result, the average angle of impact is 90 degrees for each wall, and therefore the average collision time is given by looking at 90 degree impacts for each component of the velocity vector.