So why is this a valid proof for all resistors in parallel
First, you have an error in your question - the equivalent resistance is
$$R_P = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$
Now, the voltage across the two parallel resistors is what it is regardless of how the voltage comes to be.
However we choose to label that voltage is immaterial, thus, we can arbitrarily label the voltage across the parallel resistors as, e.g., \$v_P\$.
Now, and again, it does not matter how this voltage comes to be, the voltage variable \$v_P\$ is the voltage measured across the parallel resistors when "red" lead is placed on the "\$+\$" labelled terminal and the "black" lead is on the "\$-\$" labelled terminal.
Thus, by Ohm's law, the current through each resistor is
$$i_{R_1} = \frac{v_P}{R_1} $$
$$i_{R_2} = \frac{v_P}{R_2} $$
So, the total current is, by KCL,
$$i_P = i_{R_1} + i_{R_2}$$
and the equivalent resistance is defined as
$$R_P = \frac{v_P}{i_P}$$
thus,
$$R_P = \frac{v_P}{i_{R_1} + i_{R_2}} = \frac{v_P}{\frac{v_P}{R_1} + \frac{v_P}{R_2}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$
Again, if we replace the two parallel resistors with a resistor of resistance \$R_P\$, the current through the equivalent resistance will be identical to the sum of the currents through the two parallel resistors.
Your car analogy is almost there, but not quite.
Instead of a single length of road, imagine instead a racetrack.
That racetrack is packed with cars, end to end, wall to wall. No space between them.
Now there are some narrow points on the racetrack. Each narrow point is a resistor. Each car is an electron.
The cars have to queue up to get through a narrow point. Not just because it's a narrow point, but because there are cars already in there, and cars filling the next section of road queueing to get into the next narrow point. That's the crucial difference with your analogy - you're assuming the "wire" after the resistor, and before the next resistor, is empty, but it isn't, it's full.
So the more "resistors" you have the more cars will be queuing, and the bigger the tailbacks will be.
Best Answer
Except when one is being precise, that is a fairly good explanation.
[It sometimes is the case that the work done is different depending upon path. In particular, when there is a time-varying magnetic field involved. We will get to that. But first:]
Faraday's Law states:
$$\nabla \times E = -\frac{1}{c}\frac{\partial B}{\partial t}$$
If there is no time-varying magnetic field, this implies that
$$\nabla \times E = 0$$
which means that E is a conservative field. Being a conservative field implies that E is the gradient of a potential function, (which we call V)
$$E = \nabla V$$
V, being a potential function implies that the work done to move an electron from one point to another does not depend upon the path taken. That is why the voltage across two resistors in parallel are said to be equal. And, this is true whenever there is no time-varying magnetic field present.
[However, if there is a time varying magnetic field affecting the circuit, one needs to model this as if there were a transformer, or voltage source, present in one or both of the paths through the resistors. The voltages induced by these transformers will affect the voltage "dropped" by each of the resistors.]