Consider ordinals defined as transitive sets of transitive sets with $\in$-ordering, and consider the statement in the title.
I don't know yet that $\in$ is total, but I already know that any two elements of an ordinal are $\in$-comparable.
So, the intuition is that $\alpha \in \beta \Rightarrow \alpha \cong [0; \alpha)_\beta$, in which case the claim follows in one direction. By definition, $[0; \alpha)_\beta$ is the set of elements in $\beta$ that are also in $\alpha$. It's obvious that $\alpha \subseteq [0; \alpha)_\beta$ (because if $\exists a \in \alpha, a \notin [0; \alpha)_\beta$, then $a \notin \beta$, but by definition of $\beta$: $a \in \alpha \in \beta \Rightarrow a \in \beta$), and it's also obvious $[0; \alpha)_\beta \subseteq \alpha$ (similar reasoning applies). So, by extensionality $\alpha = [0; \alpha)_\beta$ and surely $=$ implies $\cong$.
Let's now go in the other direction. Let $\alpha \cong [0; a)_\beta$ for some $a \in \beta$. Consider some larger ordinal $\gamma : \alpha, \beta \in \gamma$ (such an ordinal exists since we can construct it by transitively pulling all the required stuff from $\alpha$ and $\beta$). Now we know that either $\alpha = \beta$, or $\alpha \in \beta$ or $\beta \in \alpha$. $\alpha = \beta$ is not the case since no well-founded set can be isomorphic to a proper initial segment of itself. $\beta \in \alpha$ is not the case either since, using the first part, it would be isomorphic to some proper initial segment of $\alpha$, so we again get some set isomorphic to some proper initial segment of itself. So, we're only left with $\alpha \in \beta$, as required.
Is this reasoning correct? I feel like I've cut a corner or two somewhere.
This question (or parts of it) seems to have been asked a couple of times, but my proof seems to be different, hence I'm asking it (again).