Equivalent definition of ordinal sum I'm trying to prove that two definitions of the ordinal sum are equivalent.
In the  present question I want to prove that given
$$\tag{1}
\alpha+\beta:=\operatorname{ord}(\{0\}\times \alpha\cup \{1\}\times \beta)
$$
it follows that
$$
\alpha+ \operatorname{S}(\beta)=\operatorname{S}(\alpha+\beta)
\text{, where }\operatorname{S}(A):=A\cup\{A\}.
$$
Here is my start:
\begin{align}
\alpha+ \operatorname{S}(\beta)&=\alpha+ (\beta\cup \{\beta\})\\
&=\operatorname{ord}\big(\left(\{0\}\times \alpha\right)\cup \left(\{1\}\times (\beta\cup \{\beta\})\right)\big)\\
&=\operatorname{ord}\big(\left(\{0\}\times \alpha\right)\cup \left(\{1\}\times \beta\right)\cup \left(\{1\}\times \{\beta\}\right)\big)\\
&=\cdots\tag{*}\\
&=\operatorname{ord}\big(\left(\{0\}\times \alpha\right)\cup \left(\{1\}\times \beta\right)\big)\cup \left\{\operatorname{ord}\big(\left(\{0\}\times \alpha\right)\cup \left(\{1\}\times \beta\right)\big)\right\}\\
&=(\alpha+\beta)\cup\{\alpha+\beta\}=\operatorname{S}(\alpha+\beta)
\end{align}
I proved the result
$$
\operatorname{ord}(A\cup B)=\operatorname{ord} A+\operatorname{ord} B,
$$
but $A,B$ have to be woset and disjoint, additionally $A\cup B$ has to be woset.

How do I fill the gap? Are there already error?
  Perhaps there are easier ways (transfinite induction?), but I would like to know if there is some technical trick to fill (*).

EDIT
I changed my mind... any hint to any type of proof showing the equivalence of (1) with the following definition (2) is welcomed (incl. ref duplicate / ref. textbook):


*

*$\alpha+0=\alpha$

*$\alpha+ \operatorname{S}(\beta)=\operatorname{S}(\alpha+\beta)$

*if $\beta$ is a limit ordinal then $\alpha+\beta$ is the limit of the $\alpha+\delta$ for all $\delta<\beta$.

 A: We want to prove the following

Theorem: For each ordinal $\alpha$, the following equalities characterize the operation $\xi\longmapsto\alpha+\xi$:
$$\begin{cases}
  \alpha+0=\alpha\\
  \alpha+S(\beta)=S(\alpha+\beta)\\
  \alpha+\gamma=\text{sup}\{\alpha+\delta\;|\;\delta<\gamma\}\quad\text{if }\gamma\text{ is a limit ordinal}
\end{cases}$$

But first, let's prove the following:

Lemma: For any ordinals $\alpha,\beta$ and $\gamma$:
(a) $\alpha+0=0+\alpha=\alpha$
(b) $\alpha+1=S(\alpha)$
(c) $\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma$ (associativity)
(d) $\alpha+S(\beta)=S(\alpha+\beta)$

Demonstration:
(a) Let $\langle A,<_R\rangle$ be a well ordered set of ordinal $\alpha$. The only well ordered set of ordinal $0$ is the empty set $\varnothing$, so the ordered sum has as base set $A\cup\varnothing=A$ and its well-order is simply the well-order $<_R$ of $A$, so its ordinal is $\alpha$ and $\alpha+0=\alpha$. Similarly, we can obtain $0+\alpha=\alpha$.
(b) Let $\langle A,<_R\rangle$ be a well ordered set of ordinal $\alpha$ and let $\langle\{b\},<_S\rangle$ be a well ordered set of ordinal $1$ such that $b\not\in A$. The ordinal $\alpha+1$ of the ordered sum $\langle A\cup\{b\},<_{R\oplus S}\rangle$ is $S(\alpha)$, because
$$\langle A\cup\{b\},<_{R\oplus S}\rangle\cong\langle S(\alpha),\in_{S(\alpha)}\rangle$$
Via the isomorphism $f:A\cup\{b\}\longrightarrow S(\alpha)$ defined by: for each $a\in A\cup\{b\}$
$$f(a)=\begin{cases}
  g(a)\quad\text{if }a\in A\\
  \alpha\qquad\,\text{if }a=b
\end{cases}$$
Where $g$ is the isomorphism between the well ordered set $\langle A,<_R\rangle$ and $\alpha$.
(c) Let $\langle A,<_R\rangle,\;\langle B,<_S\rangle$ and $\langle C,<_T\rangle$ be well ordered sets of ordinals $\alpha,\beta$ and $\gamma$ respectively, such that $A,B$ and $C$ are pairwise disjoint. The ordinal $\alpha+(\beta+\gamma)$ is the ordinal of the well ordered set $\langle A\cup(B\cup C),<_{R\oplus(S\oplus T)}\rangle$, and the ordinal $(\alpha+\beta)+\gamma$ is the ordinal of the well ordered set $\langle(A\cup B)\cup C,<_{(R\oplus S)\oplus T}\rangle$. This well ordered set are not only isomorphic but identical, because $A\cup(B\cup C)=(A\cup B)\cup C$ and $<_{R\oplus(S\oplus T)}=<_{(R\oplus S)\oplus T}$, so their respective ordinals are equal, i.e. $\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma$
(d) $\alpha+S(\beta)=\alpha+(\beta+1)=(\alpha+\beta)+1=S(\alpha+\beta)$
For the last part of the theorem, we just have to prove the following:

Lemma: For any ordinals $\alpha$ and $\gamma$, if $\gamma$ is a limit ordinal, then
$$\alpha+\gamma=\text{sup}\{\alpha+\delta\;|\;\delta<\gamma\}$$

Demonstration: We must prove the two corresponding inequalities. First, for each $\delta<\gamma$, we have that $\alpha+\delta<\alpha+\gamma$ (cf. Hrbacek & Jech Introduction to Set Theory, page 120, lemma 5.4 (a)), so
$$\text{sup}\{\alpha+\delta\;|\;\delta<\gamma\}\leq\alpha+\gamma$$
On the other hand, let $\eta<\alpha+\gamma$. We shall prove that there exists some $\delta<\gamma$ such that $\eta\leq\alpha+\delta$. We have that $\eta<\alpha$ or $\alpha\leq\eta$. In the first case, we simply take $\delta=0$. In the second case, we know there exists some $\beta$ such that $\eta=\alpha+\beta$ (cf. Hrbacek & Jech Introduction to Set Theory, page 121, lemma 5.5), and since $\eta<\alpha+\gamma$, it results that $\beta<\gamma$. So taking $\delta=\beta$, we have that $\eta\leq\alpha+\delta$, and
$$\alpha+\gamma\leq\text{sup}\{\alpha+\delta\;|\;\delta<\gamma\}$$
For the other part of the equivalence, that is, supposing that we define the sum of ordinals recursively, and we want to prove that the sum of ordinals is the ordinal of the ordered sum of two well ordered disjoint sets, just take a look again at the book of Hrbacek & Jech, page 120, theorem 5.3, where this fact is proved in full detail.
A: For short let $X=(\{0\}\times \alpha)\cup (\{1\}\times \beta).$
From the 3rd line of your "start" we have $$\alpha + S(\beta)=$$ $$=ord \,(X\cup (\{1\}\times \{\beta\})\,)$$ $$==ord \,(\, (\{0\}\times X)\cup (\{1\}\times \{\beta\})\,)=$$ $$=ord\,(\,(\{0\}\times X)\cup (\{1\}\times \{\alpha+\beta\})\,)=$$ $$=ord\,(\,(\{0\}\times (\alpha+\beta)\,)\cup (\{1\}\times(\{\alpha+\beta\})
\,)=$$ $$=S(\alpha+\beta).$$
