Yes, where $d=gcd(a,b)$ with the Bezout-Bachet-theorem.

We are told the sets $N = \{-x \mid x \in S, x < 0\}$ and $P = \{y \mid y \in S, y > 0\}$ are both non-empty. Let $n= \min N$ and $s = \min S$ (they exist since $N, S \subseteq \mathbb{N}^*$, which is well-ordered). If $n\neq s$ then $0\neq s-n \in S$, and $|s-n| < \max(n,s)$, contradiction. Therefore $n = s$ (and so $0\in S$); denote this common value $d$. For $y \in P$ we have $y\geq d$, thus by Euclid's algorithm $y=qd + r$, with $0\leq r < d$. But then $r = y-qd = y+q(-d) \in S$, so $r=0$. Similar argument for $x\in N$. Therefore $S \subseteq d\mathbb{Z}$. The converse is trivial.

what about if $ 1 $ is in $ S $?

then $S=\mathbb Z$. what's the problem??

oh yes.....sorry