I'm not exactly sure if I'm understanding the question correctly but I will try and give an explanation based on my understanding.
We certainly can define $\bigcap$ as you have nicely shown. I don't think you even need the axiom of union since the intersection is in every set in $x$. Now the reason it can be defined this way means we don't need to have it as an axiom (just pointing that out in case there was confusion).
The reason we don't (usually) define the intersection in the way we do for union (i.e. as an intersection of a single set, understood as a family of sets, as it is done for union) is IMO that it's rarely used and also precisely because it's not an axiom.
Writing $\bigcap_{x\in X}x$ is easy enough and we could do it for union aswell. But the notation used for the axiom is a little neater.
But we fairly rarely take intersection of whole families, or at least I can't really think of many situations we do so.
Another reason why we do have the $\bigcup x$ notation for union is that there is parts of set theory where it does help. That is when we create transitive closures of different kinds. Essentially sometimes we are trying to make sure we have a nice big enough set where everything we need to do can be done. That is that the set we work in contains everything we could possibly need to talk about has nice properties and is still a Set as opposed to a proper class.
Your reasoning is correct in that if you define $\{x\}$ in this way, you can show that $y \in z$.
However, consider the set $z’ = \{w \in z \mid \forall k (x \in k \to w \in k\}$. Of course, $z’$ is only defined up to “equality”, and its definition requires separation.
Note that we can conclude $x \in z’$. But concluding that $y \in z’$ amounts to assuming extensionality, which we cannot do.
It’s fairly simple to come up with a model of set theory in which extensionality is false. Let $2 = \{0, 1\}$. Simply take the smallest class $U$ such that for all sets $s \subseteq U$ and all $x \in 2$, we have $(s, x) \in U$. Proving such a class exists requires some cleverness, but it’s not too hard.
Then define $\in_U$ by $w \in_U (s, x)$ iff $w \in s$. You can verify that all axioms except extensionality hold in $U$. This is an instructive model for understanding why extensionality fails; play around with it a bit, and can find an $x, y, z’$ as described above where $x$ and $y$ have the same elements and $x \in z’$ but $y \notin z’$.
Best Answer
This is a good question. Although there is something missing. When you state the Axiom of Extensionality, you are stating it as an “if and only if” condition. This axioms actually states that $$\forall x \forall y ((\forall z (z \in x \leftrightarrow z \in y)) \rightarrow x = y)\,.$$ The converse of this axiom, which is also true, is a particular instance of a standard theorem in any first order theory with equality (in this case, we consider the $\in$ as our relational symbol).
Putting both together we in fact obtain that $$\forall x \forall y ((\forall z (z \in x \leftrightarrow z \in y)) \leftrightarrow x = y)\,.$$
But this is not the definition of equality of sets. Rather, this is a characterization of the primitive notion of equality of sets.
Equality of sets is itself a primate notion (in this case a priority relation between sets), i.e. we don’t define it. Just in the same way that we don’t define the notion of set.