I've started thinking about a particular "equation-preserving" quotient of the free product of groups (or more generally, coproduct of appropriate algebraic structures). Let $\mathcal{G},\mathcal{H}$ be groups with disjoint underlying sets $G,H$ respectively. Say that a pair of $G\sqcup H$-words $w(g_1,...,g_m,h_1,...,h_n)$ and $v(g_1,...,g_m,h_1,...,h_n)$ are linked iff we have both $$\mathcal{G}\models \forall x_1,...,x_n[w(g_1,...,g_m, x_1,...,x_n)=v(g_1,...,g_m, x_1,...,x_n)]$$ and $$\mathcal{H}\models\forall x_1,...,x_m[w(x_1,...,x_m, h_1,...,h_n)=v(x_1,...,x_m,h_1,...,h_n).$$
For example, if $g\in Z(\mathcal{G})$ and $h\in Z(\mathcal{H})$ then $gh$ and $hg$ are linked. Linkage gives rise to a congruence $\sim$ on the free product $\mathcal{G}*\mathcal{H}$, namely the transitive closure of the relation $\sim_0$ given by $x\sim_0y$ iff there are linked words $w,v$ which evaluate to $x,y$ respectively in $\mathcal{G}*\mathcal{H}$. Let $\mathcal{G}\star \mathcal{H}=\mathcal{G}*\mathcal{H}/\sim$.
Intuitively, the group $\mathcal{G}\star \mathcal{H}$ connects similar "regions" of equational behavior in $\mathcal{G}$ and $\mathcal{H}$. For example, there is a canonical embedding of $Z(\mathcal{G})\times Z(\mathcal{H})$ into $\mathcal{G}\star\mathcal{H}$, but this isn't true for the free product or even the free product "modded out by" the common equational theory of $\mathcal{G}$ and $\mathcal{H}$.
Question: Does this construction have a name?
Note that we can generalize this construction to arbitrary algebras (in the sense of universal algebra) in the same signature; really the only tweak needed is to replace "word" with "term" appropriately. I've phrased this question for groups since I think it's most likely to be known in this setting, but I'm really interested in the universal algebraic version.
Too long for a comment.
First, a standard definition: Let $F = F(S)$ be a free group and $G$ a group. By the universal properties of free groups and free products, every map $S \to G$ extends uniquely to a homomorphism $F * G \to G$ that restricts to the identity on $G$. A mixed identity on $G$ is an element $w \in F*G$ such that for any map $S \to G$ the corresponding homomorphism $F*G \to G$ kills $w$. The set of mixed identities on $G$ is a normal subgroup of $F*G$, since it is by definition an intersection of kernels.
Now is this a fair interpretation of linkage? Let's assume we have presentations $F(A) \to G$ and $F(B) \to H$ (your comment suggests this is accurate with $A = G$ and $B = H$). Then we have a canonical corresponding presentation $F(A) * F(B) \to G * H$ that factors naturally through both $F(A) * H$ and $G * F(B)$. An element of $F(A) * F(B)$ is linked to $1$ if its image in $F(A) * H$ is a mixed identity on $H$ and its image in $G * F(B)$ is a mixed identity on $G$. Being an intersection of preimages of normal subgroups, the set $L$ of words linked to $1$ is a normal subgroup of $F(A) * F(B)$. Your group $G \star H$ is the quotient of $G*H$ by the image of $L$. (In particular the transitive closure is not necessary.)
Maybe some universal property argument shows that this construction doesn't really depend on the presentation?
Example:$\def\ab{\mathrm{ab}}$ Suppose $G$ and $H$ are abelian. Let $F(A)^\ab$ denote the abelianization of $F(A)$ (the free abelian group). Now for any map $B \to G$ the corresponding maps $F(A) * F(B) \to G * F(B) \to G$ factors through the abelianizations $F(A) * F(B) \to F(A)^\ab \times F(B)^\ab \to G \times F(B)^\ab \to G$. An element of $G \times F(B)^\ab$ defines a mixed identity on $G$ iff its $G$ component is trivial and its $F(B)^\ab$ component is a law (= mixed identity without constants). Similarly for $H$. Therefore the subgroup $L \le F = F(A) * F(B)$ is given by $L = (K_G \cap F(A)^f) (K_H \cap F(B)^e) F'$, where $e$ is the exponent of $G$, $f$ is the exponent of $H$, $K_G$ is the kernel of $F(A) \to G$, and $K_H$ is the kernel of $F(B) \to H$. The image of $L$ in $G*H$ is just $(G*H)'$, so $G \star H \cong G \times H$, as expected.
Challenge: Compute $S_3 \star C_2$ or $S_3 \star S_3$.