Sure, this works. In fact, one might note that, at least as the axioms are listed here, your statement is equivalent to the conjunction of the 1st, 2nd and 5th axioms which read:
- $1$ is a number.
- If $n$ is a number, then $\sigma(n)$ is a number.
- If $S\subset \mathbb N$ is a set, $1\in S$, and $n\in S \Rightarrow \sigma(n)\in S$ then $S=\mathbb N$.
One might notice that the first two axioms are the converse of the fifth - that is, the converse of the fifth is that
If $S=\mathbb N$, then $1\in S$ and $n\in S\Rightarrow \sigma(n)\in S$ and $S\subset\mathbb N$.
Given that we have both the statement and its converse, we can naturally write:
$S=\mathbb N$ if and only if $S\subset\mathbb N$, $1\in S$, and $n\in S\Rightarrow\sigma(n)\in S$.
which is precisely the same as the three axioms listed.