$\mathbb Z/p\mathbb Z=A\cup(A-A)$? $\newcommand{\Z}{\mathbb Z/p\mathbb Z}$
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.
 A: It seemed to me that the commenters know much more about this problem than they write in the comments. For that reason alone I am posting this answer, which is most likely a long comment.
For brevity, we call a symmetric set $A\subset G$ ($G$ is an abelian group) a partitioning set if $G=A\cup(A+A)$ and $A\cap(A+A)=\varnothing$.
The definition from the question (if it were formulated there) is equivalent to this since $A-A$ is symmetric, and then $A$ is also symmetric.
And we also call two partitioning sets $A$ and $B$ of $G$ equivalent if $B=\alpha(A)$ for some $\alpha\in\operatorname{Aut}(G)$.
Everywhere below $G=\mathbb{Z}_p$ for some simple $p$.
The conjecture in one of the last comments can be phrased like this:
If $A\subset\mathbb{Z}_p$ is a partitioning set, then some equivalent of it lies in
$\{-p'd,\ldots,-d,d,\ldots,p'd\}$ where $p'=[p/4]$ and $d$ is a nonzero group element.
For example, the Peter Mueller set for $p=29$ is equivalent to $[4,7]\cup[-7,-4]$
and the set for $p=13$ is equivalent to $[2,3]\cup[-3,-2]$.
Partitioning sets for some other prime:
$p=11$, $[2,3]\cup[-3,-2]$;
$p=17$ and $p=19$, $[3,5]\cup[-5,-3]$;
$p=23$, $[4,7]\cup[-7,-4]$.
In fact, the following statement is true.

Let $p>7$ be a prime. If $p=8m\pm1$ or $p=8m+3$ or $p=8m+5$ and
$P=[m+1,2m+1]$, then $A=P\cup-P$ is a partitioning set in
$\mathbb{Z}_p$.

And one more remark.
For $p=23$ we have a partitioning set $A=[-7,-4]\cup[4,7]$. This set is equivalent to the set $B=[-7,-5,-3,-1,1,3,5,7]$, which is an arithmetic progression.
A: This is a previous comment which was moved to chat by Ben Webster: In fact for every prime $p$ if $A=[-(2m-1),-m] \cup [m, 2m-1]$ for $(p+3)/8\le m<(p+3)/6$, then $\mathbb Z/p\mathbb Z$ is a disjoint union of $A$ and $A-A$, and in addition this set $A$ is not an arithmetic progression if and only if $m\ne(p+1)/6$. There are many more examples though. One peculiar sporadic one happens for $p=41$ with $A$ the multiplicative subgroup of order $10$ of $(\mathbb Z/p\mathbb Z)^\star$.
