Distribution of the Difference of Uniformly Random Elements
Firstly, why is this true?

This is easy to see, if we consider the finite field as a finite group with the addition operation (and ignore the multiplication operation)
If we consider the value $X - Y$, where $X, Y$ are uniformly and independently distributed elements of the group, then there are $n^2$ equiprobable $X, Y$ pairs possible (each with a probability $n^{-2}$.
For any group value $Z$, we have $Z = X - Y$ for $n$ possible $X, Y$ pairs (if we select any value for $X$, we see that the unique value of $Y$ that them sum to $Z$ is $X-Z$); each possible pair has probability $n^{-2}$, and so the probability for the sum to be $Z$ is $n^{-1}$.  This holds for all group members, and so the sum is equidistributed.

Secondly, what happens when you relax the algebraic conditions on the sample space?

As long as the relaxation still leaves us with a ring (that is, the addition operation is still a group), the above argument still holds.