$\newcommand{\Z}{\mathbb Z/p\mathbb Z}$ Can one partition a group of prime order as $A\cup(A-A)$ where $A$ is a subset of the group, $A-A$ is the set of all differences $a'-a''$ with $a',a''\in A$, and the union is disjoint?

As stated, the answer is "yes", at least if the order of the group is $p\equiv 2\pmod 3$, in which case one can take $A$ to be an appropriately located interval of an appropriate length: namely, $A=[n,2n-1]$ where $n=(p+1)/3$. One also can dilate $A$ replacing intervals with arithmetic progressions.

Are there any other examples where $A$ and $A-A$ partition the whole group?

Suppose that $A\cup(A-A)$ is a partition of a group of prime order; does it follow that $A$ is an arithmetic progression?

**Added two days later.** A set found by Peter Mueller in the comments can be generalized to produce infinitely many counterexamples, essentially answering the original question. Specifically, for a prime $p\equiv 5\pmod 8$ let $m:=(p+3)/8$ and define $A:=[-(2m-1),-m]\cup[m,2m-1]$. It is easily verified that then $A-A=[-(m-1),m-1]\cup[2m,p-2m]$, so that $A-A$ is disjoint from $A$, and $A\cup(A-A)$ is the whole group, while $A$ is not an arithmetic progression.