Does there exist a positive-only distribution such that the difference of two independent samples from this distribution is normally distributed? If so, does it have a simple form?
The answer to the question is No, and it follows from a famous characterization of normal distributions.
Suppose that $X$ and $Y$ are independent random variables. Then so are $X$ and $-Y$ independent random variables, and of course we can write $X-Y$ as $X + (-Y)$, the sum of two independent random variables. Now, according to a theorem conjectured by P. Lévy and proved by H. Cramér (see Feller, Chapter XV.8, Theorem 1),
If $X$ and $Y$ are independent random variables and $X+Y$ is normally distributed, then both $X$ and $Y$ are normally distributed.
The OP asks whether there exist i.i.d. positive random variables $X$ and $Y$ such that $X-Y$ is normally distributed. But even if we dispense with positivity and identical distributions, and keep only the independence, normality of $X-Y = X + (-Y)$ requires that both $X$ and $-Y$ be normal random variables. As Feller says, "the normal distribution cannot be decomposed except in the trivial manner."