I'm having trouble proving the following two ordinal multiplication properties.
If $\alpha, \beta$, and $\gamma$ are such that $\alpha \lt \beta$ and $\gamma \gt 0$, then $\alpha\gamma \le \beta\gamma$.
If $\alpha, \beta$, and $\gamma$ are such that $\gamma\alpha \lt \gamma\beta$ and $\gamma \gt 0$, then $\alpha \lt \beta$.
If you have already proved that inequality of ordinals is trichotomic, then the second part, which has the form of implication $$\gamma \alpha < \gamma \beta \qquad\Rightarrow\qquad \alpha<\beta$$ can be equivalently reformulated as $$\beta\le\alpha \qquad\Rightarrow\qquad \gamma\beta\le\gamma\alpha.$$
So now the two parts have very similar form: We have to prove for $\gamma>0$ that
(I have reversed the notation in the second parts so that the assumption in both parts is similar.)
By definition $\alpha\le\beta$ means that the ordinal $\alpha$ is isomorphic to an initial segment of the ordinal $\beta$. But it can be shown that this is equivalent to the fact that $\alpha$ is isomorphic to a subset of $\beta$. (See for example Order-isomorphic with a subset iff order-isomorphic with an initial segment. This fact was also mentioned in an answer to another question of yours).
So now we know that $\alpha\le\beta$ and we wonder whether $\gamma\alpha$ can be realized as a subset of $\gamma\beta$; and similarly for $\alpha\gamma$ and $\beta\gamma$.
This is not very difficult: The ordinal $\gamma\alpha$ simply means that we have replaced each point of $\alpha$ by a copy of the ordinal $\gamma$. This can be clearly embedded into "$\beta$-many" copies of $\gamma$.
Similarly, $\alpha\gamma$ means that we have "$\gamma$-many" copies of $\alpha$. If we take "$\gamma$-many" copies of $\beta$, we can embed each copy of $\alpha$ into a copy of $\beta$.
The above is a rather informal argument, I assume you would be able to make it more formal and describe the embeddings between the well-ordered sets we are working with.