Every proof that induction implies well ordering I have seen goes: assume $S\subset\mathbb{N}$ has no least element and let $T$ be its complement with respect to $\mathbb{N}$. Since $1$ is the smallest natural number, $1\not\in S$, so $1\in T$. If $1, 2, ..., n\in T$, then $n+1\in T$ since $1, 2, ..., n$ are the only natural numbers smaller than $n+1$. By induction, $T=\mathbb{N}$, so $S$ is empty.
Why is $1$ the smallest natural number? Is it assumed? I know well ordering implies that $1$ is the smallest natural number: if the set $S$ of all natural numbers less than $1$ is nonempty, it has a least element $x$. $x<1$, so $x^2<x$. $x^2\in\mathbb{N}$ since $\mathbb{N}$ is closed under multiplication, contradicting the minimality of $x$.
The axioms I am used to working with are: $\mathbb{Z}$ is a commutative ring under addition and multiplication, $\mathbb{N}$ is a nonempty subset of $\mathbb{Z}$ closed under addition and multiplication, trichotomy, and well ordering. I think you also need that negative times negative is positive and negative times positive is negative. If you swap induction for well ordering, can you prove that $1$ is the smallest natural number or prove well ordering a different way?