The problem you describe has definitely been considered (I remember discussing it in grad school, and at the time already it had been discussed long before then), though I can't point to any particular references in the literature. Possibly because it is linearly equivalent to uncolored graph isomorphism, as follows (this is true even for canonical forms). Call the problem you describe EQ-GI.
GI is just the special case of EQ-GI where each graph has just one equivalence class consisting of all vertices.
In the other direction, to reduce EQ-GI to GI, let $(G, \sim_G)$ be a graph with equivalence relation with $n$ vertices, $m$ edges, and $c$ equivalence classes. Construct a graph $G'$ whose vertex set consists of the vertices of $G$, together with new vertices $v_1, \dotsc, v_c$, one for each equivalence class in $=_G$, as well as $n+c+1$ new vertices $w_0, \dotsc, w_{n+c}$. Connect the $w_i$'s in a path $w_0 - w_1 - w_2 - \dotsb - w_{n+c}$, connect each $v_i$ to $w_0$, and for every vertex in $G$, connect it to the corresponding equivalence class vertex $v_i$. Then $G'$ has at most $n + 2c + n +1 \leq O(n)$ vertices and can be constructed in essentially the same time bound. (It also has at most $m + n + c + (n+c+1) \leq m + 4n + 1 \leq O(m+n)$ edges - which is $O(m)$ for connected graphs - but that's somewhat less relevant since most GI algorithms have running times that essentially only depend on $n$.)
Update: Since there was some confusion in the comments, I'm adding here a sketch of the correctness of the above argument. Given $(G_1, \sim_1)$ and $(G_2, \sim_2)$, let $G_1'$ and $G_2'$ be the graphs constructed as above; let $v_{i,1}$ denote the vertex $v_i$ from above in $G_1'$, and $v_{i,2}$ the one in $G_2'$, and similarly for $w_{i,1}$ and $w_{i,2}$. If there is an isomorphism $G_1' \cong G_2'$, it must send $w_{i,1}$ to $w_{i,2}$ for all $i$, since in each graph $w_{n+c}$ is the unique vertex that is the endpoint of any path of length at least $n+c+1$. In particular, $w_{0,1}$ maps to $w_{0,2}$. Since the neighbors of $w_0$ that aren't $w_1$ are exactly the $v_i$, the isomorphism must map the set $\{v_{1,1},\dotsc,v_{c,1}\}$ to the set $\{v_{1,2},\dotsc,v_{c,2}\}$ (and in particular both $\sim_1$ and $\sim_2$ must have the same number, $c$, of equivalence classes). Note that the isomorphism need not send $v_{i,1}$ to $v_{i,2}$ for all $i$, but is allowed to permute the indices of the $v$'s so long as the corresponding equivalence classes can be mapped to one another. Conversely, based on this description of how isomorphisms between $G_1'$ and $G_2'$ can look, it is easy to see that if $(G_1, \sim_1) \cong (G_2, \sim_2)$ then this gives an isomorphism $G_1' \cong G_2'$.
