The best explanation I can think of for you is that on the negative side of the circuit, electrons "pile up", due to the restriction of flow from the resistor. On the positive side of the terminal, there are fewer electrons (the missing electrons were "pumped" to the negative terminal), so it has an effective positive charge. When an electron crosses a resistor, it is moving from a location where similar charges are crammed together--high potential--to a location where it is balanced by opposite charges--low potential. Hence the potential drop across a resistor. It's really the buildup of charge that causes the potential across the resistor.
One important point though, is that the charges distribute themselves equally along the conductive metal wires. The electrons aren't so much attracted to or repelled from the terminals themselves, but the charges on the wires attached to the terminals. That's why the electrons don't have to make it all the way to the positive terminal to have "felt" the potential drop. Any conductive wire attached to a terminal will be at the same potential as the terminal, because the charges spread themselves out along the conductors.