It’s an immediate consequence of the fact that the positive integers are well-ordered by the usual order $<$. Let $B=\Bbb N\setminus S$; if $B\ne\varnothing$, let $b=\min B$, which exists because $<$ well-orders $\Bbb Z^+$. We proved that $1\in S$, so $1\notin B$, and therefore $b\ne 1$. Thus, $b=n+1$ for some $n\in\Bbb Z^+$. Clearly $n<b$, so $n\notin B$; but $n\in\Bbb Z^+$, so $n\in S$, and therefore by hypothesis $b=n+1\in S$, contradicting the choice of $b$. Thus, $B=\varnothing$, and $S=\Bbb Z^+$.
In the standard set-theoretic framework (ZF(C) the fact that $\Bbb Z^+$ is well-ordered by $<$ is largely a matter of definition: it’s defined to be a subset (or order-isomorphic to a subset) of $\omega$, the first infinite ordinal, and all ordinals are by construction well-ordered by $<$.
In the framework of the Peano axioms matters are a bit different: in that setting it’s actually an axiom.
But it will be true in any reasonable axiomatic setting for elementary number theory, because any such setting must match our intuitive notion of $\Bbb Z^+$, which is of a well-ordered set in which every element can be reached from $1$ in a finite number of steps.