Consider the arithmetical theory sometimes denoted by $\mathsf{R}$. The non-logical vocabulary of $\mathsf{R}$ consists of '$0$', '$S$', '$+$' and '$\times$'. The axioms of this theory are all true instances of the following schemata:
$\overline{m}+\overline{n}=\overline{k}$
$\overline{m}\times \overline{n}=\overline{k}$
$\overline{m}\neq \overline{k}$
$\forall x\,(x\leq \overline{n}\rightarrow x=\overline{0}\vee\ldots\vee x=\overline {n})$
$\forall x\, (x\leq \overline{n}\vee\overline{n}\leq x)$
where $\overline{m}$ is the standard numeral for $m$ (i.e. '$0$' preceded by $m$-many occurrences of '$S$') and $x\leq y$ is defined in the standard way e.g. $\exists z\,(x+z=y)$.
Let $\mathsf{R}^+$ be the theory that results from adding all instances of (first- or second-order) induction to $\mathsf{R}$.
There are some quantified statements that we can prove now, e.g. $\forall x\,(x\neq 0\rightarrow \exists y\,(x=S(y)))$.
However, I'd assume that there are many general claims that still can't be proved, e.g. $\forall x\forall y\,(x+y=y+x)$. Indeed, I’d assume that one cannot even prove the injectivity of the successor function.
Question: Assuming this is correct, how can I produce a (simple) model for $\mathsf{R}^+$ + ''Successor is not injective''? What do such models look like, i.e. how do you define the successor function on the domain of such models?
Specifically, I wonder what such models look like if we embed $\mathsf{R}$ into a deduction system for "second-order logic", i.e. if we add quantifiers binding predicate places and comprehension axioms.