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. Consequently, $|A_0-A_0|\le|A-A|-2\le 2|A_0|-1$, and by the induction hypothesis, $A_0$ is an arithmetic progression. Moreover, if $a_n$ were not the next term of this progression, then $(A-A)\setminus (A_0-A_0)$ would additionally contain the elements $\pm(a_n-a_2)$, leading to $|A-A|\ge|A_0-A_0|+4>2|A|-1$, a contradiction.

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Here is yet another proof; as far as simplicity is concerned (see **Noam's** answer), this one is difficult to beat. 

Suppose that $A=\{a_1,\dotsc,a_n\}$ has the property in question, where $a_1<\dotsb <a_n$. Since $a_2-a_1$ is an element of $A$ smaller than $a_2$, we must have $a_2-a_1=a_1$; that is, $a_2=2a_1$. Similarly, $a_3-a_1$ is an element of $A$ exceeding $a_2-a_1=a_1$, but smaller than $a_3$; hence, $a_3-a_1=a_2$, implying $a_3=3a_1$. Continuing this way, we get $a_k=ka_1$ for each $k=1,2,\dotsc,n$.