Consider $\omega_2 \times \omega$ and $\omega \times \omega_2$ with ordinal arithmetic.
Then $| \omega_2 \times \omega | =\omega_2$
and $| \omega \times \omega_2 | =\omega_2$
Does this imply that neither $\omega_2 \times \omega$ nor $\omega \times \omega_2$ are cardinals??
Recall the inductive definitions of ordinal multiplication:
$$\begin{align} &\alpha\cdot0=0\\ &\alpha\cdot(\beta+1)=\alpha\cdot\beta+\alpha\\ &\alpha\cdot\delta=\sup\{\alpha\cdot\gamma\mid\gamma<\delta\},\ \delta\text{ is a limit ordinal} \end{align}$$
Since $\omega$ and $\omega_2$ are certainly limit ordinals, we have that: $\omega_2\cdot\omega=\sup\{\omega_2\cdot n\mid n<\omega\}$, and since $\alpha\cdot2=\alpha+\alpha>\alpha$, whenever $\alpha>0$, we have that indeed $\omega_2\cdot\omega>\omega_2$ as ordinals.
On the other hand, $\omega\cdot\omega_2=\sup\{\omega\cdot\gamma\mid\gamma<\omega_2\}$. This is a supremum of $\omega_2$ ordinals, and we can show that each of those (of the $\omega\cdot\gamma$'s) has size $\leq\aleph_1$. Therefore $\omega\cdot\omega_2=\omega_2$, which is in fact a cardinal.