In a DC circuit with one battery and one resistor, why does the voltage drop remain the same irrespective of the value of the resistor?
"why doesn't the value of the resistance affect the potential
  difference."



this is because you consider the ideal voltage source, which, by definition, always supplies 10V (in your example) no matter which load is there, or how "hard" electrons do their work scattering over phonons in the resistor lattice. To account for "hard work" of electrons and a drop of potential, consider a more realistic model, which includes internal resistance of a battery and essentially models the hard work of electrons in realistic power source/battery.