By the definition of multiplication and addition of ordinals, the following rules follow-
Multiplication- $n\cdot\omega=\omega$ while $\omega\cdot n > \omega$
Addition- $n+\omega=\omega$ while $\omega + n >\omega$
Distribution- if $\alpha$, $\beta$ and $\gamma$ are ordinals, $\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma$ but $(\beta+\gamma)\alpha\neq\beta\alpha+\gamma\alpha$ .
Having written those, I can bring certain ordinals to a polynomial form, this way I can compare them.
For example, $(2+\omega)(\omega+3)=\omega(\omega+3)=\omega^2+\omega\cdot 3$
But what do I do with $(\omega+2)(\omega+3)$? How do I bring this to a polynomial form? The first step I am allowed to do by the rules above is:
$(\omega+2)(\omega+3)=(\omega+2)\cdot\omega+(\omega+2)\cdot3$. I can't think of a way to simplfy it from here.
Any ideas?
If those are really the only rules you're given, then I don't see how to do it.
However, remember that $\alpha\cdot\beta$ is (intuitively) the ordinal gotten by replacing each element of $\beta$ by a copy of $\alpha$. So, e.g., $\omega\cdot 2=\omega+\omega$. So, for example, we have $$(\omega+2)\cdot 3=(\omega+2)+(\omega+2)+(\omega+2),$$ which can be simplified to $\omega\cdot 3+2$. $(\omega+2)\cdot\omega$ can be handled similarly.
EDIT: in fact, we can show that those rules aren't sufficient: consider the "silly ordinal product," $\circ$, given by $\alpha\circ\beta=\alpha\cdot \beta$ if $\alpha$ is finite, and $\alpha\circ\beta=\alpha\cdot\beta+1$ if $\alpha$ is infinite and $\beta\not=0$, and $\alpha\circ 0=0$. Then $\circ$ satisfies all the rules you've written, but of course gives a different value for e.g. $\omega\cdot 2$.