This depends, as Steven Stadnicki said, on whether we assume the axiom of choice.

First, a quick observation: it's not hard to prove by transfinite induction that if $\alpha$ is an infinite ordinal, then $\vert\mathbb{N}\times\alpha\vert=\vert\alpha\vert$. *Interestingly, more is true: for all infinite ordinals $\alpha$ we have $\vert\alpha\times\alpha\vert=\vert\alpha\vert$. This is harder to prove, but not much harder.*

**If we have the axiom of choice**, then we're done: given $A$, the axiom of choice lets us prove that there is some ordinal in bijection with $A$; now just apply the previous paragraph.

**If choice fails**, however, things can get weird: it's consistent with ZF (= the usual axioms of set theory, but without choice) that there are very weird infinite sets! For example, there might be **amorphous sets** - these are infinite sets which can't be partitioned into two infinite disjoint subsets! It's easy to show that $\mathbb{N}\times A$ can't be amorphous, regardless of what $A$ is *(HINT: think about evens vs. odds ...)*.

A side comment: if choice fails, then "$\vert A\times A\vert=\vert A\vert$" must fail for some infinite set $A$: the statement "for all infinite sets $A$, $\vert A\times A\vert=\vert A\vert$" is equivalent to the axiom of choice (this was proved by Zermelo). Meanwhile, the statement "$\vert A\times\mathbb{N}\vert=\vert A\vert$" is **strictly weaker** than the axiom of choice, although it isn't provable in ZF alone.