Why does voltage drop across resistance contradict Ohm's law?
The problem with your test setup is that you are not measuring the voltage across the resistor you are swapping out. You are measuring the drop across the multimeter's internal resistance. You have a finite multimeter resistance. I'm guessing its about 1MΩ. If you compute for the resistive divider of 100kΩ and 1MΩ, you get 4.54V, roughly what you measured.
If you want to measure current, you need to either use an ammeter or a small resistor's voltage drop.