It is well known that for ordinals $\alpha, \beta, \gamma$ one has that
$$\alpha\cdot(\beta+\gamma)=\alpha \cdot \beta + \alpha \cdot \gamma$$
To show this, let $B$, $C$ be disjoint sets with respective order types $β, γ$ and $A$ a set of order-type $α$. The two orderings of $A × (B ∪ C)$ agree, namely $A×(B∪C)=(A×B)∪(A×C)$
However, doesn't the same thing hold on the left, namely $$(B∪C)×A=(B×A)∪(C×A)$$
since I feel like the exact same argument would work.
But that then implies $$(\beta+\gamma)\cdot\alpha= \beta \cdot \alpha + \alpha \cdot \gamma$$
which is false.
Where does my argument go wrong?
The problem is that neither ordinal addition nor ordinal multiplication is commutative.
The definition of $\alpha\cdot\beta$ is the order type of $\beta\times\alpha$ in the lexicographic order. It is true that $(B\cup C)\times A$ is isomorphic to $(B\times A)\cup(C\times A)$. But it is no longer necessarily the case that $(A\times B)\cup (A\times C)$ is isomorphic to $A\times (B\cup C)$.
To understand why, note that the lexicographic ordering on $X\times Y$ is given by replacing each point $x\in X$ by a copy of $Y$. This action is not commutative. Replacing each point in $\omega$ by two points, will again be $\omega$. Replacing two points by two copies of $\omega$ is not the same as $\omega$ itself.
Therefore, the obvious bijections that you have there are not order preserving. So the whole argument falls apart.