By expanding every details, I get the following:
Disproof:
We show first that possible equivalence relation for it must be unique,
1.$\ $Construct $\sim$ on $X$ such that $x\sim y$ are equivalent iff $x_k\sim_k y_k$ for each coordinate. There obviously exists bijection between both set of equivalence classes by mapping product of equivalence classes of each $\sim_i$ to equivalence class of $\sim$.
2.$\ $For uniqueness (up to homeomorphism), if there exist some $\sim'$ on $X$ and some homeomorphism between $X/_{\sim'}\ \ $ and $Y$. Consider the quotient map from $X\to X/_{\sim'}\ \ $, constant exactly on product of each equivalence class of $\sim_k$, hence on each equivalence class of $\sim$. Now by considering the quotient space $X/_{\sim}\ \ $ and the respective quotient map from $X\to X/_\sim\ \ $, obviously induces homeomorphism between $X/_\sim$ and $X/_{\sim'}\ \ $, hence this resolves to homeomorphism between $X/_\sim$ and $\prod\bigg( \ ^{X_i}/_{\sim_i}\bigg)\ \ $.
Now we just need to prove that product of these quotient maps is not a quotient map.
e.g. Let $\mathbb R^*$, $\mathbb Q^*$ be one point compactifications of $\mathbb R$ and $\mathbb Q$ respectively.
Consider $\mathbb R^*/\mathbb Z$ obtained by identiying all integers to a point, by quotient map $p$. By assuming the truth of the problem, $p\times id_{\mathbb Q^*}$ is quotient map. Since these compactifications are $T_1$, hence $\mathbb R, \mathbb Q$ are open saturated in respective space. We also have restriction of three maps to respective sets are quotient maps, denoted $q, i, q\times i$.
The map $q$ with the quotient map from $\mathbb R\to \mathbb R/\mathbb Z$ (by identifying all integers to a point), obviously induces homeomorphism between $\mathbb R/\mathbb Z\ $ and $p(\ \mathbb R) \ .$ Next, since $id_{\mathbb Q^*}$ is a homeomorphism, we have its restiction $i$ is also a homeomorphism hence $i(\mathbb Q)$ has the standard topology $\mathbb Q$. By composing with suitable homeomorphism, we can modify the codomain such that the map $q\times i$ become a quotient map between $\mathbb R\times \mathbb Q$ and $\mathbb R/\mathbb Z\times \mathbb Q$. But it is known that $q\times i$ cannot be a quotient map. Contradiction. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $(see https://i.sstatic.net/0IAjI.png)
The formulation should be more precise: if $f: X \to Y$ is surjective and continuous define $R_f$ on $X$ by $x R_f x' \iff f(x)=f(y)$, the equivalence relation induced by $f$, and let the canonical quotient map be denoted $q_f: X \to X{/}f$, sending $x$ to its class $[x]$ under $R_f$.
The function $\bar{f}: X{/}f \to Y$ (defined by $\bar{f}([x])=f(x)$ obviously ) is then well-defined set-theoretically and a bijection.
The fact the OP refers to:
$f$ is a quotient map iff $\bar{f}: X{/}f \to Y$ is a homeomorphism.
So no arbitary $h$ is involved at all.
The left to right implication is quite easy: $\bar{f}$ is continuous as $\bar{f} \circ q_f = f$ and $f$ is continuous (using the universal property for the quotient topology induced by $q_f$) and if $O \subseteq X{/}f$ is open then $\bar{f}[O] = f[O]$ which is open in $Y$ as $f^{-1}[f[O]] = q_f^{-1}[O]$ is open in $X$ and $f$ is assumed to be quotient, so that $\bar{f}$ is open and we already know from set-theory it's a bijection.
The reverse implication is quite similar. Only "$O \subseteq Y$ and $f^{-1}[O]$ open in $X$ implies $O$ open" needs to be shown. And this is clear as $O=\bar{f}[\bar{f}^{-1}[O]]$ and $\bar{f}$ is open.
Best Answer
There is a continuous bijection $$\phi : (X \times Y)/ ((X \times B) \cup (A \times Y)) \to X/A \wedge Y/B .$$ To see this, consider the the map $$\psi : X \times Y \stackrel{p_X \times p_Y}{\longrightarrow} X/A \times Y/B \stackrel{q}{\longrightarrow} X/A \wedge Y/B $$ where $p_X, p_Y, q$ are the obvious quotient maps. It is easy to verify that the fibers $\psi^{-1}(\xi)$ are singletons if $\xi \ne *$ (where $*$ is the point obtained by collapsing $X/A \times \{*\} \cup \{*\} \times Y/B$) and $\psi^{-1}(*) = X \times B \cup A \times Y$. Hence we get $\phi$.
Now $\phi$ is a homeomorphism if and only if $\psi$ is a quotient map. The problem is that the product of quotients maps $p_X \times p_Y : X \times Y \to X/A \times Y/B $ is not always a quotient map. Make a search in this forum for "product quotient maps". There are many hits, for example When is the product of two quotient maps a quotient map?
Thus you will need additional assumptions on the pairs $(X,A), (Y,B)$.
Update:
The fibers of $q : X/A \times Y/B \to X/A \wedge Y/B = (X/A \times Y/B)/(X/A \times \{*\} \cup \{*\} \times Y/B)$ are singletons for $\xi \ne *$ and $q^{-1}(*) = X/A \times \{*\} \cup \{*\} \times Y/B$.
Clearly $$(p_X \times p_Y)(X \times B \cup A \times Y) = (p_X \times p_Y)(X \times B) \cup (p_X \times p_Y)(A \times Y) \\= p_X(X) \times p_Y(B) \cup p_X(A) \times p_Y(Y) = X/A \times \{*\} \cup \{*\} \times Y/B .$$ Moreover $$(X \times Y) \setminus (X \times B \cup A \times Y) = ((X \times Y) \setminus (X \times B)) \cap (X \times Y) \setminus (A \times Y)) \\ = X \times (Y \setminus B) \cap (X \setminus A) \times Y = (X \setminus A) \times (Y \setminus B) .$$
But $p_X \times p_Y$ is injective on $(X \setminus A) \times (Y \setminus B)$ and for $x \in X \setminus A$ resp. $y \in Y \setminus B$ we have $p_X(x) \ne *$ resp. $p_Y(y) \ne *$. Thus $$(p_X \times p_Y)((X \times Y) \setminus (X \times B \cup A \times Y)) \subset (X/A \times Y/B) \setminus (X \times * \cup * \times Y/B).$$ We conclude that $(p_X \times p_Y)^{-1}(X/A \times \{*\} \cup \{*\} \times Y/B) = X \times B \cup A \times Y$ and that for $\eta \notin X/A \times \{*\} \cup \{*\} \times Y/B$ the fiber $(p_X \times p_Y)^{-1}(\eta)$ is a singleton.
This shows $\psi^{-1}(*) = X \times B \cup A \times Y$. For $\xi \ne *$ the fiber $q^{-1}(\xi)$ a singleton outside of $X/A \times \{*\} \cup \{*\} \times Y/B$, thus $\psi^{-1}(\xi)$ is a singleton.