Are all models of Peano arithmetic elementary equivalent? By Löwenheim-Skolem we know there are models of (first order) PA that are not isomorphic to the standard model, but are elementary equivalent to it, i.e. they satisfy the same set of first-order sentences.
By Gödel's incompleteness result, we know there is a model of PA in which the canonical unprovable Gödel sentence G is true (the standard model), and (non-standard) models where G is false. 
Since G is a first-order sentence, this seems to answer the question I posed in the title to the negative. Correct so far?
I am wondering then, is it a meaningful, i.e. well defined question to ask if for any "other" first-order sentences, those that intuitively express more naturally "arithmetic truths", all models of PA are elementary equivalent? are there other more arithmetically meaningful$^1$ (first order) sentences that are independent from the PA axioms?
$^1$ where 'more ... meaningful' is probably too vague to hope for an answer, but I am essentially asking, can we be sure that any first-order property we hold as evident of $\mathbb{N}$ is provable from the axioms of PA?
 A: As you say, no, not all models of $\sf PA$ are elementarily equivalent. Here's a nice point:

If $T$ is a theory without finite models, then $T$ is complete if and only if all of its models are elementarily equivalent.

So the incompleteness is a quick way to deduce that $\sf PA$ is incomplete (note that there are no finite models of $\sf PA$ because of the axioms about the successor function being injective but not surjective).

As for "normal statements", that depends on what you mean exactly. We can say that as far as $\Delta_1$ statements, the answer is positive.
The reason is that a $\Sigma_1$ statement is true in $\Bbb N$ if and only it $\sf PA$ proves it. And therefore given a $\Delta_1$ statement, both it and its negation are $\Sigma_1$, one of them is true in $\Bbb N$ and therefore provable from $\sf PA$.
Whether or not $\Delta_1$ statements are "reasonable" is up to you. And pushing it further down the arithmetical hierarchy is impossible, since $\operatorname{Con}(\sf PA)$ is a $\Pi_1$ statement and its negation is $\Sigma_1$ statement; neither of which is provable from $\sf PA$.
