Boolean-valued models are usually used for forcing. We fix a complete Boolean algebra $B$ and inductively define $V^B_0=\emptyset$, $V^B_{\alpha+1}=$ all partial functions from $V^B_\alpha$ to $B$, and take union at limit stage. The Boolean values of the formulas $u\in v$ and $u=v$ are defined inductively by
$\displaystyle|u\in v|=\bigvee_{y\in\text{dom}(v)}v(y)\land |u=y|$
$\displaystyle |u=v|=\left(\bigwedge_{x\in\text{dom}(u)}u(x)\Rightarrow|x\in v|\right)\land\left(\bigwedge_{y\in\text{dom}(v)}v(y)\Rightarrow|y\in u|\right)$
My question is: does the usual proof go through to show that $V^B$ is a Boolean-valued model of set theory if we change $V^B_0$ to some Boolean-valued structure $(M,|\cdot\in\cdot|,|\cdot=\cdot|)$ that satisfies axiom of extensionality; recall that "Boolean-valued structure" means it also has to satisfy inequalities such as $|x=y|\land|y\in z|\leq|x\in z|$. More specifically, I wonder if the following two variants produce models of $\mathsf{ZFA}$ ($\mathsf{ZF}$ with atoms) and $\mathsf{ZF^-}$ ($\mathsf{ZF}$ without axiom of foundation), respectively. For simplicity we may take $B=\{0,1\}$.
Let $V^B_0$ be a nonempty set $A$, define $|a\in b|=0$ for any $a,b\in A$ and $|a=b|=0$ for different $a,b$; I omitted obvious relations that must hold such as symmetry and reflexivity of $|\cdot=\cdot|$. We may need to require that $a$ doesn't contain any pair, or change the definition of $\text{dom}(a)$ in some way.
Let $V^B_0=\{a,b\}$, define $|a\in b|=|b\in a|=1$, $|a\in a|=|b\in b|=0$.
Edit: I forgot that $\mathsf{ZFA}$ is really a two-sorted theory, and its version of extensionality is any set is determined by its members, while atoms are different from sets. So we need to define something like a two-sorted Boolean-valued model, or use a unary relation to distinguish atoms from the empty set.