I was proving something else, and a lemma that I really could use would be stated as follows: given limit $\alpha, \beta$ the following holds:
$$ \sup \{ \sup \{ \tilde\alpha + \tilde\beta \mid \tilde\alpha < \alpha\} \mid \tilde\beta < \beta \} = \sup \{ \sup \{ \tilde\alpha + \tilde{\beta} \mid \tilde\beta < \beta\} \mid \tilde\alpha < \alpha \}. $$
This looks fairly easy with von Neuman ordinals (boiling down to interchanging the order of set unions), but how one would prove this if ordinals are viewed as sets of isomorphic well-ordered sets?
A proof that I have in mind is as follows.
Let $\sigma_1 = \sup \{ \sup \{ \tilde\alpha + \tilde\beta \mid \tilde\alpha < \alpha\} \mid \tilde\beta < \beta \}$ and $\sigma_2 = \sup \{ \sup \{ \tilde\alpha + \tilde\beta \mid \tilde\beta < \beta\} \mid \tilde\alpha < \alpha \}$. Let's check if $\sigma_1 \geq \sigma_2$, or, equivalently, $S_1 = \sup \{ \sup \{ \tilde A + \tilde B \mid \tilde A \prec A\} \mid \tilde B \prec B \}$ is an upper bound for $\sup \{ \sup \{ \tilde A + \tilde B \mid \tilde B \prec B\} \mid \tilde A \prec A \}$, where $A \in \alpha, B \in \beta$ and $\prec$ is the ordering induced by the relationship "to be a proper initial segment", and the equivalence follows from the definition of an ordinal. If that's not the case, there exists $S \in \sigma_2$ such that $S_1 \subset S$, or, equivalently, there exists $\tilde A \prec A$ such that $S_1 \subset \sup \{ \tilde A + \tilde B \mid \tilde B \prec B\}$, or, equivalently, there exists $\tilde A \prec A, \tilde B \prec B$ such that $S_1 \subset \tilde A + \tilde B$, but that contradicts the definition of $S_1$.
Thus, $\sigma_1 \geq \sigma_2$. Analogously $\sigma_2 \geq \sigma_1$. Thus, $\sigma_1 = \sigma_2$.
This doesn't have much to do with ordinals, and it certainly doesn't depend on how ordinals are realized as sets. In general, if $L$ is a complete lattice (a poset in which every subset has a least upper bound), then for any index sets $I,J$ and any elements $a_{i,j}\in L\ (i\in I,j\in J),$ if we set $b_i=\sup\{a_{i,j}:j\in J\}$ and $c_j=\sup\{a_{i,j}:i\in I\},$ it follows trivially from definitions that $$\sup\{a_{i,j}:i\in I,j\in J\}=\sup\{b_i:i\in I\}=\sup\{c_j:j\in J\}.$$