Assume the statement, and suppose that $1\in A$, $n\in A\to s(n)\in A$, and $A\ne\mathbb{N}$. Let $B=\mathbb{N}\setminus A$; clearly $B\ne\varnothing$, so $A\setminus s[A]\ne\varnothing$, so $B\setminus s[B]\ne\varnothing$. Fix $b\in B\setminus s[B]$.
You have as an axiom that $\mathbb{N}\setminus s[\mathbb{N}=\{1\}$, and $b\ne 1$ (since $1\in A=\mathbb{N}\setminus B$), so $b\in s[\mathbb{N}]$, i.e., $b=s(n)$ for some $n\in\mathbb{N}$. Now $b\notin s[B]$, so $n\notin B$, and therefore $n\in A$. But then the hypothesis on $A$ ensures that $s(n)\in A$, i.e., that $b=s(n)\in A\cap B=\varnothing$, which is absurd. This contradiction shows that in fact $A$ must be all of $\mathbb{N}$.
(Your book introduces the natural numbers in a somewhat unusual way, which accounts for the confusion in the answers and comments..)