Infinite models are partly classified by two theorems of $PA^{(-1,2)}$ the subtheory of $PA$ without axioms 1 or 2. Then I will describe the finite models completely.
In $PA^{(-1,2)}$ if axiom 1 fails then axiom 2 holds. For proof, express failure of axiom 1 by a constant $c$ with $S(c)=0$. Then $PA^{(-1,2)}$ proves $\forall x (c+S(x)=x)$, which implies axiom 2. In this case the numbers form a group under addition.
So if axiom 2 fails then axiom 1 holds. In that case we are in $PA^{(-2)}$ and axiom 2 fails at just one number $p$, there is an interval $[0,p]$ linearly ordered by addition, and the remaining numbers satisfy $PA^{(-1)}$ with $p$ in the role of 0.
To prove the claims in the preceding paragraph, consider the following formula $\Phi(x)$ which is meant to say successor is one-to-one up to $x$ (but so far we have not defined an order relation):
$\forall y,z,u,v (\ (y+z=x\ \&\ S(u)=S(v)=y) \rightarrow u=v)$
The usual proof of $x+y=0 \rightarrow y=0$ works in $PA^{(-2)}$ and the usual additive order relation is well defined on the set defined by $\Phi$. So $\Phi(0)$ and if axiom 2 fails there is some $p$ in $\Phi$ with successor not in $\Phi$ and the set defined by $\Phi$ is linearly ordered as an interval $[0,p]$.
Induction shows for every $x$ either $\Phi(x)$ or $\exists y (x=p+y)$ while conversely $\Phi(p+y)\rightarrow y=0$. In particular $p$ is successor to at least one number of form $p+y$.
The numbers of form $p+y$ provide an obvious interpretation of $PA^{(-1,2)}$ with $p$ in the role of 0, and this interpretation falsifies axiom 1. So it satisfies axiom 2. It interprets $PA^{(-1)}$.
Finite models of full induction are easy. Assuming no axioms for now, the list of iterated values $0,S(0),SS(0),\dots$ is finite in any finite model so the set of all the iterated values is definable (without parameters) by a finite disjunction $x=0\vee x=S(0)\vee x=SS(0)\vee \dots$ in that model. Assuming induction that set is the whole model.
That means (following Blass's comment) the $S$ series is eventually cyclic, so there are $m$ and $n > m$ with $0,S(0),\dots S^{n-1}(0)$ all distinct but $S^{n}(0)=S^m(0)$.
Now assume axioms 3--6 on $+$ and $\times$. Then all classically correct equalities between numerals (terms $S^p(0)$ for any number $p$) are provable, just as in $PA$ (but the classically false ones are not refutable). And every element of any model is named by at least one numeral of that form.
So, for any eventually cyclic $S$ series, at most one definition of $+$ and $\times$ produces a model of axioms 3--6 plus full induction. And one does, namely using standard arithmetic below $m$ and arithmetic mod $n-m$ above $m$. The extreme cases are helpful in understanding: if $m=0$ then we have arithmetic mod $n$ and a model of axioms 2--6 plus induction. If $m=n-1$ then we have arithmetic with $n-1$ as an absorbing upper bound, a model of axioms 1 and 3--6 plus induction.
To be a bit more explicit, given an $S$ series with $n$ and $m$ as above, define an equivalence relation $=_0$ by saying: $x=_0 y$ iff either $x=y$ and both are $\leq m$, or both are $\geq m$ and have $x=y$ mod $n-m$, You just have to show this is a congruence for $S,+,\times$ which is pretty direct since $S,+,\times$ are all strictly increasing in each argument, except multiplying by 0.
