I'm struggling to articulate this question, but here it goes.
For any first order theory $T$ involving a unary function symbol $S$ (intuitively, the successor function), let $I(T)$ denote the theory obtained by adding the first order induction schema (involving the unary function symbol $S$) to the theory $T$. For instance, if $T=\mathrm{PA}^-$ (by which I mean first order Peano arithmetic minus the induction schema), then $I(T)=\mathrm{PA}$ (by which I mean ordinary first order Peano arithmetic).
Now I believe that $I$ is a closure operator. (Can someone confirm this?) So in particular, $I$ is idempotent. Thus $I(\mathrm{PA})=\mathrm{PA}$.
Here's a definition. For all theories $T$ involving a unary function symbol $S$, lets call the theory $I$-closed if $I(T)=T.$ So the final statement of the preceding paragraph can be rephrased as "$\mathrm{PA}$ is $I$-closed."
Assuming what I've said so far is correct, my question is as follows. Let $\mathrm{Con}(\mathrm{PA})$ denote the statement that $\mathrm{PA}$ is consistent, formalized in the language of $\mathrm{PA}$. Does the theory $\mathrm{PA}+\mathrm{Con}(\mathrm{PA})$ have the property of being $I$-closed? In other words, does $$I(\mathrm{PA}+\mathrm{Con}(\mathrm{PA}))=\mathrm{PA}+\mathrm{Con}(\mathrm{PA})?$$
More generally, replace $\mathrm{PA}$ with an arbitrary theory $T'$ of arithmetic that is rich enough to express its own consistency statement. Is $T'$ necessarily $I$-closed?
Induction is normally defined over a set of formulas, not over a theory. It is not an operator on theories. I have never seen anyone write something like $I(PA)$.
Let's call your operator $O$. If we consider your definition, then $O(T)= I({\cal L}_{PA})+T$. $O(PA)=PA$ since $I({\cal L}_{PA})$ is already included in $PA$.
Now adding some other axioms doesn't change anything, your operator always adds the same set of formulas to the theory and that doesn't depend on the theory, even if you put the empty set there or the complete theory of $(\mathbb{N};0,S,+,\cdot;=)$.