Apologies if this is an elementary question that should have been obvious to me. I am learning about these topics very much from the perspective of an outside hobbyist, and am not a wizard of logic.
Do all non-standard models of PA add extra axioms? And if they do, doesn't that mean that the PA axioms do define a unique model in some sense? It's just that other models also satisfy those axioms (?).
As the comments indicate, there are several ways to show the existence of a nonstandard model of PA. The proof you've seen adds a new constant $c$ plus axioms $c>1$, $c>1+1$, etc. The ultraproduct construction takes a different route.
I don't know that I'd really describe the "$c>1+\ldots+1$" argument as adding new axioms, though. Technically that's true. But the new axioms are all about a brand new constant, and don't say anything about the standard integers.
Another way to prove the existence of non-standard models is to add an axiom like $\neg\text{Con(PA)}$ to PA. The resulting system is consistent by Gödel's second incompleteness theorem, and so has a model $N$. Since the standard model $\mathbb{N}$ satisfies Con(PA), $N$ must be nonstandard. That's more clearly a case of "adding a new axiom".
One says that $\mathbb{N}$ and $N$ are not elementarily equivalent, i.e., don't satisfy the sentences of L(PA) (the language of PA). (Note that L(PA) does not include the extra constant $c$.) So $\mathbb{N}$ and $N$ do not "satisfy the same axioms" in any sense.
The comments also mentioned that PA is not $\aleph_0$-categorical. This is a somewhat different (but related) issue. $\text{Th}(\mathbb{N})$ is so-called "true arithmetic" (or "full arithmetic"): the set of all sentences of L(PA) that are true for $\mathbb{N}$. (While we have no way of deciding, in general, exactly which sentences belong to $\text{Th}(\mathbb{N})$, it's still interesting theoretically.)
Both the "$c>1+\ldots+1$" argument and the ultraproduct construction show that there exists a model, call it $M$, of $\text{Th}(\mathbb{N})$ that is nonstandard. So $M$ and $\mathbb{N}$ are elementarily equivalent but not isomorphic.
We can also insist that $M$ is countable. That's what "not $\aleph_0$-categorical" means: two non-isomorphic countable models. So not just PA, but even $\text{Th}(\mathbb{N})$ is not $\aleph_0$-categorical.
So PA fails to define a unique model in two senses: there are other countable non-isomorphic models, and even stronger, there are other non-elementarily equivalent models.