Certainly the good money is on the (current) nonexistence of such an example. But that's boring. Here are a couple positive observations, although - in my opinion - each falls short of an actual positive answer, and indeed I don't know of anything I would call a "natural" nonstandard model of $\mathsf{PA}$.
First, here's a basic model-theoretic observation. Say that a structure $\mathfrak{M}$ is pointwise-definable iff every element of $\mathfrak{M}$ is (parameter-freely-)definable in $\mathfrak{M}$. Some structures are pointwise-definable (e.g. $(\mathbb{N};+,\times)$ or $(\mathbb{Q};+,\times)$) while others aren't (e.g. $(\mathbb{R};+,\times)$ or $(\mathbb{Q};+)$). Similarly, some but not all theories have pointwise-definable models, and some theories have pointwise-definable models even though we might reasonably expect them not to (such as $\mathsf{ZFC}$). Of course, whenever a complete theory does have a pointwise-definable model it has exactly one up to isomorphism (just "match up the definitions" and check that everything works).
Now for a structure $\mathfrak{A}$, let $\mathfrak{A}^{\mathsf{def}}$ be the substructure (allowing emptiness) of $\mathfrak{A}$ consisting of all definable elements (ignoring the issue that the latter might be empty). In general $\mathfrak{A}^{\mathsf{def}}$ might be very unlike $\mathfrak{A}$; however, it turns out that models of $\mathsf{PA}$ behave very nicely in this regard: we always have $\mathfrak{A}^{\mathsf{def}}\preccurlyeq\mathfrak{A}$ when $\mathfrak{A}\models\mathsf{PA}$. This is because we have full induction in $\mathsf{PA}$: think about the Tarski-Vaught test and note that the least number satisfying a formula is always definable.
Consequently every completion of $\mathsf{PA}$ has a pointwise-definable model. By the uniqueness observation above this means that arguably every natural completion of $\mathsf{PA}$ (other than true arithmetic, $\mathsf{TA}$, itself) gives rise to a natural nonstandard model of $\mathsf{PA}$. Of course we still need to come up with a "natural" completion of $\mathsf{PA}$ other than $\mathsf{TA}$, and I don't really think it's clear how to do that, but I do think this is still a positive-"flavored" result in some sense.
What about nonstandard models of $\mathsf{TA}$? The unique pointwise-definable model of $\mathsf{TA}$ is obviously just $\mathbb{N}$ itself, so the above doesn't help. Here we can turn to ultrapower constructions, and - despite the obvious obstacles - there are still a couple of positive things to be said.
First, assuming $\mathsf{CH}$ we have that any two nonprincipal ultrafilters on $\omega$ yield isomorphic ultrapowers of $\mathbb{N}$ (or indeed any other countable structure - see here). So "the nontrivial $\omega$-fold ultrapower of $\mathbb{N}$" is actually a proper description of a nonstandard model of arithmetic, and the non-canonical nature of the ultrafilter used turns out to disappear.
In $\mathsf{ZFC}$ alone a more involved construction appears: not a single ultrapower, but a way of combining all possible ultrapowers in a sense. In broad terms, Kanovei and Shelah produced a way to construct, unambiguously, a hyperreal field which works just assuming $\mathsf{ZFC}$. The natural numbers part of this field then is a nonstandard model of $\mathsf{TA}$. I don't personally consider this a "natural" nonstandard model, but it is certainly highly interesting and canonical in some sense.
OK, one last shot:
Solving all our problems by getting rid of the axiom of choice
(Wait, what in the whatting what?!)
Working in $\mathsf{ZF}$ alone, it is consistent that there are infinite cardinalities (which noun incidentally becomes a bit more subtle than in choice-land) which behave a bit like natural numbers. Specifically, Sageev showed the following:
Suppose that $\mathsf{ZFC}$ + "There is an inaccessible cardinal" is consistent. Then there is a model of $\mathsf{ZF}$ + "There is a Dedekind-finite infinite set" in which the Dedekind-finite cardinalities are linearly ordered by injectability.
This turns out to be highly relevant to our question due to earlier (and more elementary but harder to find) work by Ellentuck:
In any model of $\mathsf{ZF}$ where the Dedekind-finite cardinalities are linearly ordered by injectability, they in fact form a nonstandard model of $\mathsf{TA}$ (with "$+$" and "$\times$" coming from disjoint union and Cartesian product as expected).
Put together these results show that certain models of $\mathsf{ZF}$ come with in my opinion highly natural and surprising canonical nonstandard models of true arithmetic. Of course this requires choice to fail extremely badly and much about these models is still unclear (see 1, 2), so I still don't think this constitutes a positive answer to your question, but still - pretty dang cool, no?
Best Answer
A simple method for constructing a nonstandard model of arithmetic is to use compactness as follows. Let $c$ be a new constant symbol not occurring in the language of arithmetic and add to PA the infinitely many new axioms $0<c$, $S0<c$, $SS0<c$, etc. This new theory is consistent, since any finite subset $F$ of it contains only finitely many of the new axioms and has a model consisting of the standard model with the new constant $c$ interpreted as a sufficiently large integer to cover the finitely many new axioms occurring in $F$.
Now take a model of the (whole) new theory -- it is a model of PA and contains an element (the interpretation of $c$ ) which is bigger than all the standard natural numbers $0, S0, SS0$, etc. To be precise, the reduct of this model to the language of PA is the model you want.
Finally, this construction has nothing to do with incompleteness. You could start with complete arithmetic (i.e. all first order sentences true in the standard model) instead of PA and then we'd have proved the existence of non-standard models which satisfy exactly the same sentences of arithmetic as does the standard model.