Assuming that your sets are finite and consist of positive integers only, every such set is a homogeneous arithmetic progression $\{d,2d,\dotsc,nd\}$, where $n=|A|$.
For the proof, let $A-A$ be the set of all differences $a_1-a_2$ with $a_1,a_2\in A$, and denote by $D_+$ the set of all positive elements of $A-A$. We have $D_+\subsetneq A$ (the inclusion is strict since the largest element of $A$ is not in $D_+$) and $|D_+|=(|A-A|-1)/2$. It follows that $|A-A|=2|D_+|+1\le 2|A|-1$, which is known to be possible if $A$ is an arithmetic progression only. It is then easy to see that the progression must be of the indicated form $\{d,2d,\dotsc,nd\}$.
Addressing a question in the comments. To derive from $|A-A|=2|A|-1$ that $A$ is an arithmetic progression, one can use induction. Write $A=\{a_1,\dotsc,a_n\}$ with $a_1<\dotsb<a_n$, and let $A_0:=A\setminus\{a_n\}$. We have $(A_0-A_0)\cup\{a_n-a_1,a_1-a_n\}\subseteq A-A$, with the union in the left-hand side disjoint. Moreover, if $A_0$ is a progression (the induction hypotheses), while $A$ is not, then $(A-A)\setminus (A_0-A_0)$ additionally contains the elements $\pm(a_n-a_2)$.