Why is the voltage drop across an ideal wire zero?

The key thing is that there is NO electric field within the perfect wire. So, there is no force acting on the electron, and thus no work done on it (while it's in the perfect wire).

This goes back to the definition of a perfect conductor (which the perfect wire is). Within a perfect conductor, there is no electric field. Instead, the charges (which have infinite mobility) rearrange themselves on the surfaces of the conductor in such a way as to perfectly cancel out any internal field.

So, the only fields in your circuit would be 1) in the battery, and 2) in the resistor.

I should also add that this is due to the approximation of the wire as 'perfect'. A real wire has some resistance, or equivalently, its charges don't perfectly reorder so as to perfectly cancel an internal field.

$F$ is not greater than 0 in an ideal wire, think 'frictionless surface' if it helps. In this idealisation, the electrons are considered to move from 4 to 3 without effort... Therefore there is no need to invoke any energetic loss.

If this doesn't appeal then you need to drop the idealisation and consider resistivity and then you can the case more physical...

If you're feeling philisophical: "In an idealisation, there is no wire."