I have found a way to fix my previous issue and posted here.
Lemma 1: Let $A$ be the set of ordinals, $B\subseteq A$ such that $\forall\alpha\in A,\exists\beta\in B:\alpha\le\beta$. Then $\sup A=\sup B$.
Proof:
Show $\{\gamma \mid \forall\alpha\in A : \gamma \ge\alpha\} = \{\gamma \mid \forall \beta\in B: \gamma \ge\beta\}$ and thus the
minima of these sets, and hence the suprema of $A$ and $B$, are equal.
Suppose $\forall\alpha\in A : \gamma \ge\alpha$. Then, since $B\subseteq A, \forall \beta\in B: \gamma \ge\beta$.
Suppose $\forall \beta\in B: \gamma \ge\beta$. For any $\alpha\in A$, there is some $\beta\in B$ such that $\alpha\le\beta$, hence $\gamma \ge\beta\ge\alpha$. Thus $\gamma\ge\alpha$ for all $\alpha\in A$.
Lemma 2: $\alpha+\beta<\alpha+\gamma\iff \beta<\gamma$ for all ordinals $\alpha,\beta,\gamma$.
We proceed to prove our theorem.
Let $B:=\{\alpha+(\beta+\delta)\mid\delta<\gamma\}=\{\alpha+(\beta+\delta)\mid\beta+\delta<\beta+\gamma\}$ by Lemma 2, and $A:=\{\alpha+\epsilon\mid\epsilon<\beta+\gamma\}$. Our task is to prove $\sup A=\sup B$.
It's clear that $B\subseteq A$. By Lemma 1, our task is done if we show $\forall a\in A, \exists b\in B:a\le b$.
Let $\alpha+\epsilon\in A$ and $\delta:=\min\{\xi\mid\beta+\xi\ge\epsilon\}$.
$\delta\le\gamma$. If not, $\delta>\gamma$ and thus $\gamma>\beta+\epsilon$, which is a contradiction.
$\delta<\gamma$. If not, $\delta=\gamma$. For each $\xi<\gamma,\xi<\delta$ and thus $\beta+\xi<\epsilon$. Then $\epsilon<\beta+\gamma=\sup\limits_{\xi<\gamma}(\beta+\xi)\le\sup\limits_{\xi<\gamma}(\epsilon)=\epsilon$, which is a contradiction.
By construction, $\beta+\delta\ge\epsilon$ and thus $\alpha+\epsilon\le\alpha+(\beta+\delta)\in B$ by Lemma 2. Thus the conditions of Lemma 1 are satisfied.
Let me provide a few more details. First let's prove the stated
Lemma. Let $1 < \alpha$, let $\beta$ be a limit ordinal and let $A \subseteq \beta$ be cofinal. Then $a^\beta = \sup \{ a^\gamma \mid \gamma \in A \}$.
Proof. By the definition of ordinal exponentiation we have that $\alpha^\beta = \sup \{ \alpha^\gamma \mid \gamma < \beta \}$. Since $\{\alpha^\gamma \mid \gamma \in A \} \subseteq \{ \alpha^\gamma \mid \gamma < \beta \}$ we thus clearly have that $\sup \{ \alpha^\gamma \mid \gamma \in A \} \le \alpha^\beta$.
Conversely let $\xi < \alpha^\beta$. Since $\alpha^\beta = \sup \{ \alpha^\gamma \mid \gamma < \beta \}$ there is some $\gamma < \beta$ such that $\xi < \alpha^\gamma$. Now, since $A \subseteq \beta$ is cofinal, there is some $\gamma < \gamma^* \in A$. Since $\gamma \mapsto \alpha^\gamma$ is increasing, we have that $$\xi < \alpha^\gamma < \alpha^{\gamma^*} \le \sup \{ \alpha^{\delta} \mid \delta \in A \}.$$
Thus $\alpha^\beta = \sup \{ \alpha^\gamma \mid \gamma \in A \}$. Q.E.D.
Now to the main event:
Lemma. Let $\alpha > 1$, $\beta, \gamma$ be ordinals. Then $\alpha^{\beta + \gamma} = \alpha^{\beta} \cdot \alpha^{\gamma}$.
Proof. We proceed by induction on $\gamma$ and only consider the limit case. (As seen in the original post, the case $\gamma = \delta + 1$ is easy.)
We have
$$
\begin{align*}
\alpha^{\beta + \gamma } &= \sup\{ \alpha^ \delta \mid \delta < \beta + \gamma \} && \beta + \gamma \text{ is a limit ordinal} \\
&= \sup\{ \alpha^ \delta \mid \beta \le \delta < \beta + \gamma \} && \text{ see lemma above} \\
&= \sup \{ \alpha^{\beta + \delta} \mid \delta < \gamma \} && \text{every } \beta \le \delta < \beta + \gamma \text{ can be written as } \delta = \beta + \delta^* \text{ for a unique } \delta^* < \gamma \\
&= \sup \{ \alpha^\beta \cdot \alpha^\delta \mid \delta < \gamma \} && \text{ induction hypothesis}
\end{align*}
$$
It remains to be seen that $\sup \{ \alpha^\beta \cdot \alpha^\delta \mid \delta < \gamma \} = \alpha^\beta \cdot \alpha^\gamma$.
Let $\delta < \gamma$. Then $\alpha^\delta < \alpha^\gamma$. Since $(\xi, \eta) \mapsto \xi \cdot \eta$ is increasing in the second coordinate, it follows that $\alpha^\beta \cdot \alpha^\delta < \alpha^\beta \cdot \alpha^\gamma$ and thus $\sup \{ \alpha^\beta \cdot \alpha^\delta \mid \delta < \gamma \} \le \alpha^\beta \cdot \alpha^\gamma$.
Conversely, let $\xi < \alpha^\beta \cdot \alpha^\gamma$. Since $\cdot$ is increasing in the second coordinate, there is then some $\eta < \alpha^\gamma$ such that $\xi < \alpha^\beta \cdot \eta$. Since $\alpha^\gamma = \sup \{ \alpha^\delta \mid \delta < \gamma \}$ there is moreover some $\delta < \gamma$ such that $\xi < \alpha^\beta \cdot \alpha^\delta$. This shows that $\alpha^\beta \cdot \alpha^\gamma \le \sup \{ \alpha^\beta \cdot \alpha^\delta \mid \delta < \gamma \}$ and hence the desired equality. Q.E.D.
Best Answer
This comes from a more general fact:
Note that by definition $\sup A=\bigcup A$. Let $\beta:=\sup A$.
If $\beta=0$ then $(1)$ holds trivially.
If $\beta=\delta+1$, you can show that necessarely $\beta\in A$, i.e, $\beta=\max A$. Therefore $(1)$ holds.
If $\beta$ is a limit ordinal, then by the very definition of ordinal addition, $\alpha+\beta=\sup\{\alpha+\gamma\mid\gamma\in\beta\}$, and you can easily verify that $$\sup\{\alpha+\gamma\mid\gamma\in\beta\}=\sup\{\alpha+\gamma\mid\gamma\in A\}$$ by looking at the cases where $\beta\in A$ and $\beta\notin A$.