The simplest example is $$1+\omega=\omega \neq \omega +1$$
But I think it is also true that $$\omega ^\alpha+\omega^\beta=\omega ^\beta$$ for general ordinals $\alpha < \beta$ but I don't know how to prove this.
In fact I don't even know how to prove $1+\omega^2=\omega^2$.
I've tried induction on $\beta$:
If $\beta=\sup(\lambda| \lambda <\beta)$ is a limit, then given $\alpha<\beta$, we have $$\omega^\alpha+\omega^\lambda=\omega^\lambda$$ for all $\alpha<\lambda<\beta$. Taking sups, we get $$\omega^\alpha+\omega^\beta=\sup(\omega^\alpha+\omega^\lambda)=\sup(\omega^\lambda)=\omega^\beta$$
If $\beta=\hat{\beta}^+ $ is a successor, and $\alpha<\hat{\beta}$ then $$\omega^\alpha+\omega^{\hat{\beta}}=\omega^{\hat{\beta}}$$ implies $$\omega^{\alpha^+}+\omega^\beta=\omega^\beta$$ because ordinal multiplication is left-distributive. This still leaves $\alpha=0$ and $\alpha=\hat{\beta}$ which I don't know how to do.
You can prove in general that if $\beta \ge \alpha\cdot \omega$, then $\alpha+\beta=\beta$.
The easiest way is probably to write $\beta=\alpha\cdot\omega+\delta$, and then construct ordered sets of order type $\alpha+\alpha\cdot\omega+\delta$ and $\alpha\cdot\omega+\delta$ and show an order isomorphism between them. The $\delta$ parts maps to each other; mapping $\alpha+\alpha\cdot\omega$ to $\alpha\cdot\omega$ can be done by increasing the second entry in each pair in $\alpha\cdot\omega$ by one and map the $\alpha$ part to $\alpha\times\{0\}$.
Expressing this with purely syntactic induction arguments looks like it's going to be complex, but nobody says you have to do that if you have shown once and for all that the order-type interpretations of ordinal arithmetic work.