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

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

$\beta < \gamma \ \Longrightarrow \ \beta + \alpha < \gamma + \alpha \qquad \text{for all ordinals } \alpha, \beta, \gamma.$

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 $0 \ne 1$.

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.