What's missing so far is a rigorous proof that $1$ is not even. Here is one that starts from the Peano arithmetic.
Suppose $1=S(0)$ is even. That means that for some $a$ we have that $S(0)=a+a$. Now either $a=0$ or $a=S(b)$ for some $b$. (This can be proved by induction on $a$ if you don't already know it).
In the case where $a=0$, we have $S(0)=0+0=0$, but $S(x)\ne 0$ for all $x$ by axiom, so this is a contradiction.
On the other hand, if $a=S(b)$, then $$S(0)=S(b)+S(b)=S(S(b)+b)=S(b+S(b))=S(S(b+b))$$
where in the middle step I'm using commutativity of addition which I assume we have already proved. But the successor function is required to be injective, so the extremes of this equation gives $0=S(b+b)$ which again contradicts the $S(x)\ne 0$ axiom.
Therefore there is no $a$ such that $1=a+a$ -- in other words, $1$ is not even.