Let $A$ be a coalgebra over a field $\mathbb K$ and $\mathcal A = \mathrm{Comod}^A$ its category of comodules. A well-known result of Sweedler writes $A = \mathrm{colim}_i A_i$ where $A_i$ are finite-dimensional subcoalgebras of $A$. Let $A_i^*$ denote the linear dual to $A_i$ and $\mathcal A_i = \mathrm{Mod}_{A_i^*} = \mathrm{Comod}^{A_i}$ the category of $A_i^*$-modules, equivalently the category of $A_i$-comodules. As observed in my paper with Brandenburg and Chirvasitu, $\mathcal A = \mathrm{colim} \mathcal A_i$ is the colimit (over the same diagram) in the bicategory of $\mathbb K$-linear locally presentable categories. Here and throughout I sloppily write "$=$" for "there is a canonical equivalence".
Tensor products of coalgebras distribute over colimits of coalgebras, and tensor products of categories distribute over colimits of categories. Moreover, it is well-known that if $R$ and $S$ are $\mathbb K$-algebras, then $\mathrm{Mod}_{R \otimes S} = \mathrm{Mod}_R\boxtimes \mathrm{Mod}_S$. (This can be checked directly from the universal property of the tensor product.)
Suppose that $B = \mathrm{colim}_j B_j$ is another coalgebra over $\mathbb K$ with $B_j$ its finite-dimensional subcoalgebras, and $\mathcal B = \mathrm{Comod}^B$ and $\mathcal B_j = \mathrm{Mod}_{B_j^*}$ as above. We have
$$ \begin{aligned}\mathcal A \boxtimes \mathcal B & = (\mathrm{colim}_{i} \mathcal A_i) \boxtimes (\mathrm{colim}_j \mathcal B_j) \\ & = \mathrm{colim}_{i,j} (\mathcal A_i \boxtimes \mathcal B_j) \\ & = \mathrm{colim}_{i,j} (\mathrm{Mod}_{A_i}^* \boxtimes \mathrm{Mod}_{B_j^*}) \\ & = \mathrm{colim}_{i,j} (\mathrm{Mod}_{A_i^* \otimes B_j^*}) \\ & = \mathrm{colim}_{i,j} (\mathrm{Comod}^{A_i \otimes B_j}) \\ & = \mathrm{Comod}^{\mathrm{colim}_{i,j} A_i \otimes B_j} \\ & = \mathrm{Comod}^{A\otimes B}. \end{aligned}$$
On the other hand, the tensor product of categories can be presented as follows. For arbitrary $\mathbb K$-linear locally presentable categories $\mathcal A$, $\mathcal B$, objects of $\mathcal A \boxtimes \mathcal B$ are colimits over objects of the form $V \boxtimes W$ subject to the relation that $\mathrm{colim}_m (V_m \boxtimes W) = (\mathrm{colim}_m V_m) \boxtimes W$, and similarly in the $W$-variable, where on the LHS the colimit is computed in $\mathcal A \boxtimes \mathcal B$ and on the RHS the colimit is computed in $\mathcal A$.
It follows from this presentation that if $\mathcal A$ and $\mathcal B$ are both semisimple categories, then so is $\mathcal A\boxtimes \mathcal B$. Indeed, suppose that $\mathcal A$ is semisimple and let $\mathcal I = \{I_1,\dots\}$ be a complete set of simples. By "semisimple" I mean that every object of $\mathcal A$ is a direct sum of objects from $\mathcal I$. Let $\mathbb K_m = \mathrm{End}(I_m)$; by Schur's lemma, it is a division ring. Then $\mathcal A = \bigoplus_m \mathrm{Mod}_{\mathbb K_m}$. If $\mathcal B$ is also semisimple with simples $\{J_1,\dots\}$ and $\mathrm{End}(J_n) = \mathbb L_n$, then
$$ \mathcal A \boxtimes \mathcal B = \bigoplus_{m,n} \mathrm{Mod}_{\mathbb K_m} \boxtimes \mathrm{Mod}_{\mathbb L_n} = \bigoplus_{m,n} \mathrm{Mod}_{\mathbb K_m \boxtimes \mathbb L_n}.$$
Suppose now that $\mathbb K$ is algebraically closed and $\mathcal A = \mathrm{Comod}^A$ and $\mathcal B = \mathrm{Comod}^B$. Then $\mathbb K_m = \mathbb L_n = \mathbb K$ for all $m,n$. It follows that every object of $\mathcal A \boxtimes \mathcal B$ is a direct sum of objects of the form $I_m \boxtimes J_n$, as you wanted. (Under the equivalence $\mathcal A \boxtimes \mathcal B = \mathrm{Comod}^{A\otimes B}$, $I \boxtimes J$ corresponds to the $A\otimes B$-comodule $I \otimes J$.)
If $\mathbb K$ is not algebraically closed, then one can find many examples where $\mathbb K_m \boxtimes \mathbb L_n$ is not a division ring. It is always, however, a semisimple algebra, and so its module theory is semisimple with simples given by direct summands of the regular module. Thus in this case every object of $\mathrm{Comod}^{A\otimes B}$ is a direct sum of direct summands of objects of the form $I_m \boxtimes J_n$.
So as you can see, the answer to your question correctly interpreted is "yes", and it has nothing to do with Hopf algebras or quantum groups.
