I think I have the solution to this in my head, It is to interpret in $\text{ZF-Ext.}$ (with the last version of replacement) the theory $\text{ZFA}$ which in turn interpret $\text{ZF}$.
The crux of the proof is to use Marcel Crabbe` approach to interpreting equality and membership used in the equalization of $\text{SF}$ to $\text{NFU}$. So equality will be interpreted as co-extensionaity, and membership as membership in SETs, where SETs are defined as unions of equivalence classes under coextensionality, denote those as $``=^*"$ , $``\in^*"$, now we can see that the "atomic" formulas of that language (i.e. whose formulas only use $``=^*"$ , $``\in^*"$ as predicate symbols), are preserved under co-extensionality, i.e. if truth value of $x \in^*\alpha$ is K (for whatever $x$ and $\alpha$), then if we replace $x$ by a co-extensional object to $x$ and if we replace $\alpha$ by a coextensional object to $\alpha$ still the resulting formula would have truth value K, and of course the same applies to atomic formula $ x=^* \alpha$, no doubt, this to be called as "truth preservation under co-extensional replacements".This means that for any formula in the language that only uses $``=^*"$ , $``\in^*"$ as predicates, then its truth is preserved under co-extensional replacements, because the truth of any formula is a function of the truth of its atomic subformulas, so if the truth of latter ones is not affected by such replacements then the truth of the total formula shall not be affected. Call this LEMMA.
Now important theorems is that this theory proves that for any $A$ we can have a set of all objects coextensional with elements of $A$, simply let $\phi(x,z)$ in the replacement scheme to be $z \in x$, and $B$ would be a set of all $y's$ that are coextensional with the $x's$ in $A$, Call this PRPOSITION.
Also $\in$-Separation (using any kind of formulas in the language of set theory) is provable in an easy manner, simply let $\phi(x,z)$ in the replacement scheme to be $z=x \wedge \pi(x)$, and take the union of the resulting set from Replacement, and we get the set of all $x \in A$ satisfying $\pi$.
By the above LEMMA if the separating formula $\pi$ in $\in$-Separation only uses $=^*, \in^*$, then clearly the asserted object is a SET (i.e. a union of equivalence classes under coextensionality).
So $\in^*$-Separation is PROVED for ($=^*, \in^*$ formulas).
The proof of Weak Extensionality, Pairing, Union and Power-{0}, is rather straightforward (all written using predicates $=^*$,$\in^*$, over the whole domain of $\text{ZF-Ext.}$).
Now the proof of replacement is pretty much straightforward also since if the replacement formula satisfies the condition $\exists k \forall y (\phi(x,y) \iff y=^*k)$, then matters are really easy, i.e. we are replacing each $x \in A$ by every $y$ that satifies $\phi(x,y)$ and all of these $y$'s are coextensional! now to do that we take $\phi(x,z)$ , then apply replacement and each $y$ will be the set of all coextensional objects to a $z$ that fulfills $\phi(x,z)$, since there must exist such a $z$ by the preconditioning on replacement, then by pairing there must exist a set $\{z\}$ where $z$ fulfills $\phi(x,z)$, and by PROPOSITION we'll have the set of all coextensionals of $z$, and so the union of all such sets would be the needed replacement set, which is simply the union of the resulting replacement set. This would be a SET because of LEMMA.
Of course we can easily define transitive closures in this method, and we can actually define PURE SET as a set that is hereditarily set! and we can take the domain of our model to be the class of all hereditarily sets, and the equality and membership relation defined as above over that domain. This will save us headache about the many Ur-elements and easily interpret all the hierarchy of ZF(with full Extensionality).
The axiom of collection is easily provable in the usual manner by sending elements $x$ of $A$ to all minimal stages of the hierarchy that contains a $y$ as an element such that $\phi(x,y)$, this can be easily done by using the above replacement, and thus proving even collection.
The proof of Infinity goes as follows:
What is needed is to prove that:
$\exists N [\exists o \in^* N (\not\exists z (z \in^* o) )\wedge \\ \forall x \in^*N (\forall y (\forall z (z \in^* y \iff z \in^* x \lor z =^* x) \implies y \in^*N)]$
The idea is to define the usual Von Neuman ordinals (i.e. transitive classes of transitive classes such that no two distinct elements are co-extensional) as $\in$-ordinals, and to define $\in^*$-ordinals as transitive SETs of transitive SETs (Assuming Regularity), then to define a “copying relation” between them such that for each $\in$-ordinal there is a copying relation $F$ that sends it to an $\in^*$-ordinal, and the latter shall be denoted as an $\in^*$-ordinal copy of it. We’ll denote a usual finite $\in$-ordinal by $n$ and an $\in^*$-ordinal copy of it by $n^*$. Now we define the predicate "ordinal copying relation" in the following manner:
For all $F$, $F$ is an ordinal copying relation from $n$ to $m$ if and only if $F$ is a set of Kuratowski ordered pairs such that:
$[\forall a,b (\langle a,b \rangle \in F \implies a \in n \wedge b \in m)] \wedge \\ [\forall a,b,c,d (\langle a,b \rangle \in F \wedge\langle c,d\rangle \in F \implies (a=c \iff b=^*d))] $
also $F$ must satisfy the “quasi-bijectivity” condition which is:
$[\forall a \in n \exists b \in m*(\exists p \in F (p=\langle a,b \rangle))] \wedge \\ [\forall b \in m \exists a\in n(\exists p \in F (p=\langle a,b \rangle))]$
Now we can also require "quasi-isomorphism" condition, i.e. that of:
$\forall a,b,c,d (\langle a,b \rangle \in F \wedge \langle c,d\rangle \in F \implies (a\in c \iff b\in^*d)) $.
Now all $\in^*$-empty ordinals are $\in$-empty sets, so a quasi-bijectivity ordinal copying relation trivially exists between them. So any first Von-Neumann ordinal will have an $\in^*$-ordinal copy, which is simply itself! And actually any other empty set. Now we can prove by induction that for any “finite” Von-Neuman ordinal $n$, there is an $n^*$ copy of it, and also there is a set of all $n^*$ copies of it. This is easy since this is the case for every empty set $n$, and if $n$ has an $n^*$ and a set $F$ that is an ordinal copying relation between it and that $n^*$, then clearly $n+1$ will have an $(n+1)^*$ and a set $G$ that is an ordinal copying relation between them, since $G$ will be a union of $F$ with a Cartesian product of an $\{n\} $ and an $\{n’| n’ =^* n^*\}$; and so an $(n+1)^*$ will be a range of $G$.
So in nutshell for every $\in$-ordinal $n$ that has an $\in^*$-ordinal copy $n^*$, then each successor $n+1$ of $n$ will be easily shown to be sent by some ordinal copying relation to each successor $(n+1)^*$ of $n^*$, where $(n+1)^*$ is a union set of $n^*$ and a set $\{n’| n’ =^* n^*\}$. Now any set $\{n’| n’ =^* n^*\}$ would be a set that contains ALL co-extensionals of any set $n^*$ because simply ALL $n^*$ sets are co-extensional! so this proves that for every finite von Neuman ordinal $n$, there is a set of all sets coextensional to each $\in^*$-ordinal copy of it.
Now by Replacement we can send each element $n$ of the set of all finite Von Neumann ordinals to all sets $y$ of all of $n^*$ copies. Now a union of this replacement set would be a set that corresponds to an $\in^*$-$\omega$. Or another way is to use Replacement (last version above) to replace each element $n$ of the set of all finite Von Neumann ordinals by each set $y$ of all $\in^*$-ordinal copies of elements of $n$, in other words each $n$ would by replaced by all $n^*$ sets, and so the resulting set from Replacement would be an $\in^*$-$\omega$ SET. QED
So I think this is a sketch of a proof of ZF being interpretable in ZF-Ext. with the above axiom of replacement.
