To simplify notation define $\delta(x,y) = |x-y|$. Here's an easier proof (using only Euclid) of the result in the answer that Seva posted: a finite set of positive integers closed under $\delta$ a homogeneous arithmetic progression $\{d, 2d, \ldots, nd\}$.

Let $d = \gcd(A)$. By Euclid's algorithm, $\gcd(x,y)$ can be obtained from $x,y$ by repeated application of $\delta$, so $A$ is closed under pairwise $\gcd$; hence by induction $d \in A$. Then $\max(A) = nd$ for some integer $n$, and $A \subseteq \{d, 2d, \ldots, nd\}$. By repeated application of $\delta(\cdot,d)$ we find that $A$ also contains $(n-1)d$, $(n-2)d$, $(n-3)d$ etc., so $A \supseteq \{d, 2d, \ldots, nd\}$.
  Therefore $A = \{d, 2d, \ldots, nd\}$, QED.

It soon follows that if $A$ is an infinite set of positive integers and $A$ is closed under $\delta$ then $A$ consists of all positive multiples of $\gcd(A)$.

To simplify notation define $\delta(x,y) = |x-y|$. Here's an easier proof (using only Euclid) of the result in the answer that Seva posted: a finite set of positive integers closed under $\delta$ a homogeneous arithmetic progression $\{d, 2d, \ldots, nd\}$.

Let $d = \gcd(A)$. By Euclid's algorithm, $\gcd(x,y)$ can be obtained from $x,y$ by repeated application of $\delta$, so $A$ is closed under pairwise $\gcd$; hence by induction $d \in A$. Then $\max(A) = nd$ for some integer $n$, and $A \subseteq \{d, 2d, \ldots, nd\}$. By repeated application of $\delta(\cdot,d)$ we find that $A$ also contains $(n-1)d$, $(n-2)d$, $(n-3)d$ etc., so $A \supseteq \{d, 2d, \ldots, nd\}$. Therefore $A = \{d, 2d, \ldots, nd\}$, QED.

It soon follows that if $A$ is an infinite set of positive integers and $A$ is closed under $\delta$ then $A$ consists of all positive multiples of $\gcd(A)$.