Find all $A\subseteq\mathbb{N}$ such that $A=\{|a-b|:a,b\in A\}$.

For a set $A$ of real numbers, denote $$A^\ast:=\{|a-b|:a,b\in A\}.$$

Question: Find all finite subsets $A\subseteq\mathbb{N}$ of the natural numbers such that $$A^*=A.$$

Attempt: The empty set is a trivial solution, so let's assume $A\neq\emptyset$. Clearly, one must have $0\in A$. Also, it is clear that $$A=\{0,1,2,\ldots,n\}$$ is a solution for each $n\in\mathbb{N}$. More generally, $$A=\{0,k,2k,\ldots,nk\}$$ is a solution for any $n,k\in\mathbb{N}$. I am tempted to say that these are the only possible $A$, but not sure how to prove it.

The conjecture is correct.

Suppose $A=\{0,k,\ldots, s\}$ where $k$ is the smallest positive element of $A$, and $s$ is the largest. Then $s-k\in A^\star=A$, and then $s-2k\in A$. Continuing in this way, if $s$ is NOT a multiple of $k$, then $s-\alpha k$ will be in $(0,k)$ for some $\alpha$, which contradicts the choice of $k$. Hence in fact $s$ is a multiple of $k$, and we have learned that $$\{0,k,2k,\ldots,nk\}\subseteq A$$ Suppose now there is some $m\in A$ that is not a multiple of $k$. Since $\{0,k,\ldots, nk\}$ are each $k$ apart, $m$ must land in between two of them. Hence, there is some $\alpha k\in A$ such that $0<|m-\alpha k|<k$. But then $m-\alpha k\in A$, which again contradicts the choice of $k$.