How is electromagnetic interference possible when measuring voltage difference?
Now imagine you have the world's worst EMI and it causes an extra watt to be added to the signal wire and the reference voltage wire. The difference between the wires is still two so the signal hasn't changed.


This is why differential wiring is often chosen. It does help to reject radiated interference.


  What gives? Why does EMI cause nasty noises at the other end when it would presumably be changing both wires' power equally?


It won't change both wires' power (or voltage or current) exactly equally.

If one wire is slightly further from the source than the other (even by millimeters) then it will see a slightly lower E field from that source. At high frequencies, the signal reaching the further wire will be significantly out of phase with the one reaching the closer wire.

Furthermore, any AC magnetic field in the space between the two wires will induce an EMF in one direction on one wire, and in the opposite direction on the other, exactly a differential current signal. To combat this we can twist the wires so that the EMFs induced in one section of the wire pair oppose those produced in another section. But the twist is never perfectly uniform, and the field is never exactly uniform over the length of the wire pair.

Finally, your differential receiver can not perfectly reject common mode signals, so in your example, where a extremely large common mode interference signal is injected on the wire, imperfections in the receiver will cause this to confound your attempts at a differential mode measurement.