Your question rests on a few misconceptions, but it's still a good question.
First, since electrons have negative charge, their actually pushed from the point of lower electrical potential towards a point of higher potential.
Second, we don't normally think of a single electron travelling all the way through the resistor. Many many electrons exist in the material to begin with. When an electric field is applied, they're all pushed together towards the higher potential. Some of them are free to move and so they move.
In fact, the electrons that are free to move are mostly already moving, randomly in different directions. When the field is applied, it just tends to slightly skew the distribution of their direction of motion so that the overall trend is for the electrons to be moving toward the higher potential.
But along the way they're likely to interact with atomic nuclei or other electrons and bounce around, resulting in the randomized motion we just discussed.
Each time an electron "bounces" off an atom in the material, it can give up a bit of its kinetic energy to the atom, and set it vibrating. This vibration can be transferred to the other atoms nearby, and the overall combination of different vibrations is what we experience as heat.
As for whether the electrons that come out the far end (the high potential end) of the resistor still have some kinetic energy (and electrical potential energy), yes they do. And they continue to experience resistance as they (roughly) travel down whatever wire connects them to the + terminal of the battery. But the resistance of the wire is (if the wire is chosen correctly) so small compared to the resistance of the resistor, that we can ignore it for most purposes.
Best Answer
Assuming we are talking about the circuit below:
...then the voltage across the resistor is indeterminate. If we assume the batteries are mathematically ideal voltage sources, then the circuit is like saying 5 = 1. It can't be solved because it's inconsistent to begin with.
Just because ideal batteries create a mathematical problem doesn't mean building the circuit with real batteries breaks any physics (but it will quickly break the batteries themselves, wire used to connect them, and/or the experimenter). Actual batteries and the wire used connect them have some a small but nonzero series resistance. If we model the internal resistances of the batteries then the circuit to analyze looks like this:
For a typical battery, R1 and R2 will be under an ohm. The series resistance makes the circuit solvable.