Imagine for example the following structure. It consists of the natural numbers, coloured blue, together with the integers, coloured red. The successor operation is the natural one.
If you want a more formal description, our structure $S$ is the union of the set of all ordered pairs $(a,0)$, where $a$ ranges over the natural numbers, and the set of all ordered pairs $(b,1)$, where $b$ ranges over the integers. If $x=(a,0)$, define $S(x)$ by $S(x)=(a+1,0)$, and if $x=(b,1)$, define $S(x)$ by $S(x)=(b+1,1)$.
This structure $S$ satisfies your axiom, but is quite different from the natural numbers.
There are much worse possibilities. In the above description of $S$, instead of using all pairs $(b,1)$ where $b$ ranges over the integers, we can use all pairs $(b,1)$, where $b$ ranges over the reals.
Remark: Because of the wording of the question, we addressed only the issue of order type, which is settled by the second-order version of the induction axiom, and is indeed equivalent to it. Order types of models provide only weak information about the structure of models of first-order Peano arithmetic.
There is an important set-theoretic issue here: to be "categorical", a theory must have only one "model". If some reasonable candidate for second-order ZFC was categorical, its unique "model" would have to be the class of all sets. But then it would not have a set model, so it would actually be inconsistent in second-order semantics.
Put another way: there is a cardinal $\kappa$ such that if a countable theory in second-order logic has any model, then it has a model of size less than or equal to $\kappa$ (this is related to the Löwenheim number of second-order logic). But it would not make sense to call a set theory "second order ZFC" if its unique model had size less than $\kappa$, since we know there are sets larger than $\kappa$. And no matter what countable second-order theory we consider, we will never manage to exceed $\kappa$. (Surely any reasonable candidate for second-order ZFC would have at most a countable number of axioms.) So most of the set-theoretic universe would be omitted by any candidate for "categorical second-order ZFC".
Nevertheless, there are theories that are often called "second order ZFC". One such theory is just ZFC, but with the axiom scheme of replacement replaced by a single second-order axiom that quantifies over every class function $f$, and says that the image of any set under any class function is again a set. These theories are not categorical in second order logic, but at least they are consistent, and their models are much more nicely behaved than arbitrary models of first-order ZFC.
Best Answer
There is an important distinction to make. The second-order induction axiom is just that - an axiom. It can be interpreted in several ways.
In normal second-order arithmetic, $Z_2$, we do have the usual second-order induction axiom $$ (\forall X)\big [(0 \in X \land (\forall n)[n \in X \to n+1\in X]) \to (\forall n)[n \in X]\big ] $$ But $Z_2$ is usually studied with first-order semantics, and in that context it is an effective theory of arithmetic subject to the incompleteness theorems. In particular, $Z_2$ includes every axiom of PA, and it does include the second-order induction axiom, and it is still incomplete.
Therefore, the well-known categoricity proof must not rely solely on the second-order induction axiom. It also relies on a change to an entirely different semantics, apart from the choice of axioms. It is only in the context of these special "full" semantics that PA with the second-order induction axiom becomes categorical.
Now, if we fix a sound deductive system for $Z_2$, the change of semantics has no effect whatsoever on the formulas that are provable. So, even though $Z_2$ with full second-order semantics is categorical, for any sound effective deductive system there are still true formulas of $Z_2$ that are neither provable nor disprovable in that system.
This answers a question in the comments, "How can a categorical theory be incomplete?" The answer is that categoricity is determined both by the choice of axioms and the choice of semantics, while completeness in this sense is determined by the axioms and the choice of a deductive system. (Here "complete" means that the set of provable theorems is a maximal consistent set.) There is no reason why, in a general setting, categoricity should imply completeness. In fact, it doesn't.
No matter what semantics we wish to use, it is simply impossible - by the incompleteness theorems - to come up with any effective deductive system that extends PA and is complete in the sense of the previous paragraph. Moving to higher-order systems helps us prove additional true propositions about the natural numbers, but we can never find a deductive system to prove all of them.