Solution:
If $S$ has just one element, then $S=\{a\}$ satisfies the given condition.

Let the elements of $S$ be $a_1 < a_2 < \cdots < a_n$.

If $a_1 < 0$, then $|a_n - a_1| \geq a_n - a_1 > a_n$, so $|a_n - a_1|$ cannot be an element of $S$—contradiction.

Now suppose $a_1 > 0$. The numbers
$$\begin{aligned} b_2 & = a_2 - a_1 \\ b_3 & = a_3 - a_1 \\ & \vdots \\ b_n & = a_n - a_1 \end{aligned}$$
are all positive differences of members of $S$, so each is in $S$. But each $b_k$ is less than $a_k$, and greater than $b_2, \ldots, b_{k-1}$, so by induction we must have $b_k = a_{k-1}$. Thus, $a_{k-1} = a_k - a_1$ for $k = 2, 3, \ldots, n$, which gives $a_k = k a_1$ for each $k$. Thus the members of $S$ form the arithmetic progression $a_1, 2a_1, 3a_1, \ldots, n a_1$.

Finally, if $a_1 = 0$, then we can remove $a_1$ and the remaining members of $S$ still satisfy the condition of the problem. Then, by the previous case, they are $a_2, 2a_2, \ldots, (n-1)a_2$. Hence, the members of $S$ form the arithmetic progression $0, a_2, 2a_2, \ldots, (n-1)a_2$.