I'm trying to prove or disprove the following statement.
Let $M$ and $N$ be smooth manifolds. $M$ and $N$ are parallelizable if and only if $M\times N$ is parallelizable.
To prove the direct implication, I reason as it follows:
$M$ and $N$ parallelizables implies $TM \approx M \times \mathbb{R}^m$ and $TN \approx N \times \mathbb{R}^n.$ Hence $$T(M\times N) \approx TM \times TN \approx (M \times \mathbb{R}^m) \times (N \times \mathbb{R}^n) \approx (M \times N) \times \mathbb{R}^{m+n}.$$
But I can't justify the first diffeomorphism and I have read it's canonical, how can I see that without going deep in Category Theory?
I think the converse is false. I'm trying to see that $S^2 \times \mathbb{R}$ is parallelizable, but I can't find the 3 independent vector fields.
Let $\pi_1 : M\times N \to M$ and $\pi_2 : M\times N \to N$ be projections onto the first and second factors respectively. Then $T(M\times N) \cong \pi_1^*TM\oplus\pi_2^*TN$. If $TM$ and $TN$ are trivial, then so is $T(M\times N)$. That is, if $M$ and $N$ are parallelisable, so is $M\times N$.
As you suspect, the converse is not true. To see that $S^2\times\mathbb{R}$ is parallelisable, note that it is diffeomorphic to $\mathbb{R}^3\setminus\{(0,0,0)\}$ via the map $((x, y, z), t) \mapsto (e^tx, e^t y, e^tz)$. As $\mathbb{R}^3$ is parallelisable, it follows that $S^2\times\mathbb{R}$ is also parallelisable.