FALSE: $\beta <\gamma$ implies $\beta +\alpha <\gamma +\alpha$

FALSE. Ordinal addition (Ordinal addition $\alpha +\beta$) is strictly increasing in its left argument:

$\beta <\gamma ⟹\beta +\alpha <\gamma +\alpha \text{for all ordinals }\alpha ,\beta ,\gamma \mathrm{.}$

What is true is the weak inequality $\beta \le \gamma \Rightarrow \beta +\alpha \le \gamma +\alpha$, which is claim (c) of Monotonicity of ordinal $+$ and $\cdot$: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0+\beta =\beta$ and $1\cdot \beta =\beta$. The strict version fails already at $\beta =0$, $\gamma =1$, $\alpha =\omega$, so the weak form is best possible. Right cancellation fails with it: $0+\omega =1+\omega$ with $0\mathrm{\ne }1$.

Why the left argument is the weak side. The recursion of Ordinal addition $\alpha +\beta$ runs on the right argument, and at a limit it takes a supremum; a finite head placed on the left is swallowed by that supremum. Concretely, prepending finitely many points to a copy of $\omega$ gives a copy of $\omega$ again. On the right nothing is swallowed, and there the inequality really is strict, which is claim (b) of Monotonicity of ordinal $+$ and $\cdot$: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0+\beta =\beta$ and $1\cdot \beta =\beta$.

How much can be lost on the left. As much as one likes below the limit: $n+\omega =\omega$ for every $n\in \omega$, by the same computation as step 2.1 with $1$ replaced by $n$. So the map $\beta \mapsto \beta +\omega$ is constant on $\omega$ and collapses infinitely many values.

Left cancellation is unaffected. $\alpha +\beta =\alpha +\gamma$ still forces $\beta =\gamma$, because addition is strictly increasing in the right argument. The two cancellation laws are not a package, and this item is exactly the difference.