Fiber products do commute with sequential colimits of closed embedding in CGWH, but one must remember that neither limits nor colimits in CGWH are the same as the corresponding limits and colimit in the category TOP of topological spaces. If you want to keep the fiber product and sequential colimits of TOP, here is a setting where this whould work:
Claim: Fiber products commute with sequential colimits of closed embeddings if the $X_i$'s and $Y_i$'s are Hausdorff locally compact and the $Z_i$'s are Hausdorff.
Proof:
As mentioned in the question we have a continuous bijection
$$ f:colim_i(X_i \times_{Z_i} Y_i) \longrightarrow colim_i(X_i) \times_{colim_i(Z_i)} colim_i(Y_i) $$
Our first step is to reduce to the case where $Z_i = \ast$ for every $i$. To do so consider the continuous bijection
$$ g:colim_i(X_i \times Y_i) \longrightarrow colim_i(X_i) \times colim_i(Y_i) $$
By the universal property we may identify $colim_i(X_i) \times_{colim_i(Z_i)} colim_i(Y_i)$ with a subspace of $colim_i(X_i) \times colim_i(Y_i)$ (equipped with the subspace topology). On the other hand, since each $Z_i$ is Hausdorff the maps $h_i:X_i \times_{Z_i} Y_i \longrightarrow X_i \times Y_i$ are closed embeddings. We claim that the induced map $h:colim_i(X_i \times_{Z_i} Y_i) \longrightarrow colim_i(X_i \times Y_i)$ is a closed embedding as well. First since sequential colimits of sets preserve injective maps it follows that $h$ is injective. Now let $A \subseteq colim_i(X_i \times_{Z_i} Y_i)$ be a closed set. Then the intersection of $h(A)$ with the $X_i \times Y_i$ coincides with $h_i(A \cap (X_i \times_{Z_i} Y_i))$, which is closed. Hence $h(A)$ is closed as desired. We may now conclude that the topology on $colim_i(X_i \times_{Z_i} Y_i)$ is the one induced from $colim_i(X_i \times Y_i)$. It will hence suffice to proce that the map $g$ is a homeomorphism. In other words, we may assume that $Z_i = \ast$.
Let $W$ be an open subset of $colim_i(X_i \times Y_i)$. We wish to show that $g(W)$ is open. For each $i$ let us denote by $W_i = W \cap (X_i \times Y_i)$, considered as an open subset of $X_i \times Y_i$. Let $(x,y) \in colim_i(X_i) \times colim_i(Y_i)$ be a point which belongs to $g(W)$. Then there exists an $i$ such that $x \in X_i$ and $y \in Y_i$ and we may find open neighborhood $x \in U_i \subseteq X_i$ and $y \in V_i \subseteq Y_i$ such that $U_i \times V_i \subseteq W_i$. Furthermore, since $X_i$ and $Y_i$ are locally compact we may assume in addition that the respective closures $\overline{U}_i$ and $\overline{V}_i$ are compact and that $\overline{U}_i \times \overline{V}_i \subseteq W$. Our next goal is to find open subsets $U_{i+1} \subseteq X_{i+1}$ and $V_{i+1} \subseteq Y_{i+1}$ such that $U_{i+1} \cap X_i = U_i$, $V_{i+1} \cap Y_i = V_i$, the closures $\overline{U}_{i+1}, \overline{V}_{i+1}$ are both compact, and $\overline{U}_{i+1} \times \overline{V}_{i+1} \subseteq W$. To construct $U_{i+1}$ and $V_{i+1}$ we begin by choosing for every $(u,v) \in \overline{U}_i \times \overline{V}_i$ open neighborhoods $u \in A_{u,v} \subseteq X_{i+1}$ and $v \in B_{u,v} \subseteq Y_{i+1}$ such that $\overline{A}_{u,v}$ and $\overline{B}_{u,v}$ are compact and $\overline{A}_{u,v} \times \overline{B}_{u,v} \subseteq W_{i+1}$. For each $u \in \overline{U}_i$ we may then find a finite collection $B_{u,v_1}, ...,B_{u,v_n}$ which covers $\overline{V}_i$. We now set $C_u = A_{u,v_1} \cap ... \cap A_{u,v_n}$ and $D_u = B_{u,v_1} \cup ... \cup B_{u, v_n}$. We then observe that $C_u$ is an open neighborhood of $u$ in $X_{i+1}$, $D_u$ is an open neighborhood of $\overline{V}_i$ in $Y_{i+1}$, and $\overline{C}_u \times \overline{D}_u \subseteq W$. Next we may find a finite collection $C_{u_1},...,C_{u_m}$ which covers $\overline{U}_i$. We may now set $E_{i+1} = C_{u_1} \cup ... \cup C_{u_m}$ and $F_{i+1} = D_{u_1} \cap ... \cap D_{u_m}$. Then $E_{i+1}$ is an open neighborhood of $\overline{U}_i$ in $X_{i+1}$, $F_{i+1}$ is an open neighborhood of $\overline{V}_i$ in $Y_{i+1}$, $\overline{E}_{i+1}$ and $\overline{F}_{i+1}$ are both compact and $\overline{U}_i \times \overline{V}_i \subseteq \overline{E}_{i+1} \times \overline{F}_{i+1} \subseteq W_{i+1}$. Finally, since $X_i \longrightarrow X_{i+1}$ is a closed embedding we may let $U_{i+1}$ be the intersection of $E_{i+1}$ with some open subset of $X_{i+1}$ whose intersection with $X_i$ is $U_i$. Similarly we define $V_{i+1} \subseteq Y_{i+1}$. Then $U_{i+1} \cap X_i = U_i$ and $V_{i+1} \cap X_i = V_i$. Furthermore, $\overline{U}_{i+1}$ and $\overline{V}_{i+1}$ are compact and $\overline{U}_{i+1} \times \overline{V}_{i+1} \subseteq W_{i+1}$, as desired.
To finish the proof we apply the procedure described repeatedly so that for every $j=i,i+1,...$ we have open subsets $U_j \subseteq X_j, V_j \subseteq Y_j$ such that $\overline{U}_j$ and $\overline{V}_j$ are compact and $\overline{U}_j \times \overline{V}_j \subseteq W_j$. Furthermore, by construction we have $U_{j'} \cap X_j = U_j$ and $V_{j'} \cap Y_j = V_j$ for every $j < j'$. Let $U = \cup_{j \geq i} U_j$ and $V = \cup_{j \geq i} V_j$. Then $U$ is an open subset of $colim_i X_i$, $V$ is an open subset of $colim_i Y_i$ and $(x,y) \in U \times V \subseteq g(W)$. This show that $g(W)$ is open, as desired.
