How can we tell if a set of axioms uniquely determines an algebraic structure? Up to isomorphism.
For instance, the group axioms are verified by an infinite number of non-isomorphic algebraic structures. But the Peano axioms, I think (my proof may lack some formality due to my lack of in depth knowledge of logic and foundations, but I think it's rigorous enough that someone more knowledgeable could make it into a formal proof) uniquely determine $\mathbb{N}$: any two structures that satisfy the Peano axioms must be isomorphic.
One thing I noted is that the axioms for standard classes of structures (rings, groups etc - all of which have many non-isomorphic instances) are always of the form:
$$(\forall a_1...a_n\in A)\ \phi(a_1...a_n) = \psi(a_1...a_n)$$
Where $\phi$ and $\psi$ are functions "defined in terms of" the structure operations. On the other hand, the Peano axioms include axioms of other forms, notably the ones governing the successor operations. Could this be relevant?
Are there any results describing the relationship between an axiom set, and the number of non-isomorphic structures verifying it?
 A: Actually, Peano Arithmetic does not uniquely determine $\mathbb{N}$, even if you fix the cardinality of the structure you want (also called a model). These are the so-called "non-standard" models of PA, and there are a lot of them: $2^{\aleph_0}$ to be more precise. They all look like "end-extensions" of $\mathbb{N}$. That is, $\mathbb{N}$ plus a bunch of stuff at the end. There are some things you can say about what such a model looks like and how its arithmetic behaves. Read more about that here.
That link also shows how to show that such models exist, but let me outline a construction here using Godel's incompleteness theorem. Godel proved, among other things, that there is a logical sentence $G$ that is true in $\mathbb{N}$, but cannot be proved or disproved from the axioms of PA. Thus if we consider the axioms PA $+ \neg G$, Peano Arithmetic plus the the negation of $G$, this is consistent, and hence has a model by standard results in logic. Of course, any model of PA $+ \neg G$ also satisfies PA, but this model disagrees with $\mathbb{N}$ on the truth of $G$, so they cannot be isomorphic.

To go into a little more detail about your question as a whole, let me explain what Peter Smith brought up in the comments. There is a notion of a "countably categorical" or "$\omega$ - categorical" theory. This is a collection of axioms where every countable model is isomorphic. You can think of this as uniquely determining a (countable) structure. Given a collection of axioms (which we will call a theory), it is of interest to determine when it is or is not $\omega$-categorial, how many models there are, and perhaps how much "variation" there can be. There are many equivalent conditions for $\omega$-categoricity outlined here, but I will try to give you some context before sending you to read Wikipedia.
You can think of a theory as the basic operating system behind your model. The theory tells you what axioms you must satisfy. However, there are all sorts of possible extra software packages you can install. These extra packages are called "types". Some of these types are not compatible with your operating system, so you give up on them immediately. Some software packages are compatible with your operating system, but not with each other. Each "type" promises you the existence of a collection of special objects in your model. A $1$-type gives you one special element, whereas a general $n$-type gives you a collection of $n$ elements satisfying certain properties with respect to each other, and the model in general. Unless I'm mistaken, I don't think that types are the final word on the issue of non-isomorphic models, but they are certainly important to that end. Understanding what sort of types are compatible with our theory, and in what combination, tells us a lot about the sorts of models our theory can have. 
A: If there is a sentence specifying the cardinality of the model to be a finite $n$ then you can write an axiom schema which ensures its uniqueness. 
But if a [consistent] first-order theory does not have finite models then it has models in every cardinality. This follows from the Lowenheim-Skolem theorems. Clearly, then, those models are not isomorphic. 
Sometimes, though, a theory has a unique model in a particular cardinality, this is called categoricity. There are several tests whether or not a theory is categorical in any cardinality, but those usually require some more knowledge in model theory. 
It is not a coincidence that model theory and algebra have a large intersection in modern mathematics. 
A: A footnote. The OP says in his comments that his putative proof that  any two structures that satisfy the Peano axioms must be isomorphic assumes that $0$, $S(0)$, $S(S(0))$, ... are the only elements in the structure.
Suppose $T$ is a theory of arithmetic which contains at least Robinson Arithmetic (i.e. the usual axioms for successor, and recursion axioms for addition and multiplication). Say that a model of $T$ is slim iff the only elements of its domain are the denotations of '$\mathsf{0}$', '$\mathsf{S(0)}$', '$\mathsf{S(S(0))}$'. It is easily shown that all the 'slim' models of $T$ are isomorphic. So a fortiori, the slim models of first-order PA are all isomorphic, and so are slim models of second-order PA.
So yes, IF the OP is entitled to the assumption that his version of PA only has slim models, then his conclusion the theory is categorical is correct.
BUT: if we dealing with first-order PA, then the assumption is illegitimate. There can be non-slim models that verify all the first-order axioms (as other answers have noted).
On the other hand, if we are dealing with second-order PA, then the assumption is correct. In fact, Dedekind's categoricity proof proceeds by showing that all the models of second-order PA are slim.
