The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue).
The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so there is no flow. As soon as the circuit is closed, the electrons have a path and start to flow across the conductor (a resistor is also a conductor).
The only thing the resistor will govern, is how strong the flow of electron will be from A to B (from one potential to another, but remember this potential was created by the generator, the resistor doesn't need to know it).
I often do the analogy between electricity and hydraulic flow. Consider the difference of potential (the voltage) created by the generator as a height, and the resistor as a
slope. The flow of water will be different in the same way as the electrons in a circuit.
Consider the 3 scenario in the following image:
No resistor to close the circuit => no current flow possible.
Low resistor. The electrons (or the water) fall from the high potential to the low one (rather rapidly).
C: The electrons (or the water) fall from high to low potential, still from the same height (your Voltage drop is identical). Except this time the higher resistance (flatter slope) makes it harder to reach the low potential (=> lower current).
Now imagine the
high potential is 24V, or even 10,000, the mechanics are still the same, the electrons will flow from one potential to another as soon as they have a path, regardless of the resistor value. The only difference made by the resistor is how strong they will flow (how strong the current will be).
note: The analogies with the water flow are easily arguable and quickly reach their limits with complex circuits, this is not the exercise here. They still are a great way to explain electric current in the simple cases