$\text{Mod}(A)$ is an $E_n$ category $\Leftrightarrow$ $A$ is an ??? algebra Say we're working in a symmetric monoidal $\infty$-category $\mathcal{S}$, and $A$ is an associative algebra in it. For instance,
$$\mathcal{S}\ =\ \text{dg vector spaces},\ \ \ A\ =\ \text{a dg algebra}.$$
Then:

*

*$\text{Mod}(A)=\text{Mod}_\mathcal{S}(A)$ is just a plain category, no extra structure (i.e. an $E_0$ category).

*$A$ is a bialgebra iff $\text{Mod}_\mathcal{S}(A)$ is an $E_1$ category, i.e. is monoidal, and if the forgetful functor is monoidal.

*$A$ is a quasitriangular bialgebra (e.g. quantum group) iff $\text{Mod}_\mathcal{S}(A)$ is an $E_2$ category, i.e. is braided monoidal, and if the forgetful map to $\mathcal{S}$ is a braided monoidal functor.

*(This next step is what my question is about)

Question: What about for slightly higher $n$, like $E_3,E_4,E_5,...$? Is there a simple definition of what $A$ has to be in that case? Are there examples of such $A$ coming up in real life? Quantum groups are extremely interesting and deep objects, so one imagines that the next ones along will be even more interesting.
 A: $E_0$ is not quite "no extra structure" : you know who $A$ is inside $Mod(A)$, so it's a pointed category (more generally, an $E_0$ object in $\mathcal S$ is an object with a "unit" $\mathbb 1\to X$).
For a different version of your question, the answer is "$A$ is $E_{n+1}$ iff $Mod(A)$ is $E_n$", where "iff" can be made more precise : the space of $E_n$-structures on the $E_0$-category $Mod(A)$ (i.e. the space of $E_n$-structures where $A$ is the unit) is equivalent to the space of $E_{n+1}$-structures on $A$ extending its $E_1$-structure.
But that does not seem to be the question you're asking, you seem to be asking about monoidal structures on $Mod(A)$ such that the forgetful functor $Mod(A)\to \mathcal S$ is strict monoidal - at least that's how I understand your answer for $n=1$ (you say "bialgebra", where the previous paragraph hints at $E_2$-algebra, which shows we aren't thinking of the same thing).
So what's the answer for your version of the question ? To answer this, I'll make a few observations. But before that, I'll set the stage, where probably the assumptions are not necessary but they'll make things easier: I'll assume $\mathcal S$ is presentably symmetric monoidal, so that I can work in $\mathcal S$-modules in $Pr^L$ and have a nicely behaved Lurie tensor product $\otimes_\mathcal S$.
1- The functors appearing in your structure are of the form $Mod(A)\otimes_\mathcal S Mod(A)\to Mod(A)$, over $\mathcal S \simeq \mathcal S\otimes_\mathcal S\mathcal S\to \mathcal S$, or variations thereon with more tensor products. One of the nice properties of $\otimes_\mathcal S$ is that $Mod(A)\otimes_\mathcal S Mod(B) \simeq Mod(A\otimes B)$. In particular your functors are of the form $Mod(A\otimes A)\to Mod(A)$, over $\mathcal S$. Because limits are preserved and reflected by the forgetful functors, your functors preserve both limits and colimits.
2- In particular, the full subcategory of $Mod_\mathcal S(Pr^L)_{/\mathcal S}$ on objects of the form $Mod(A)\overset{forgetful}\to \mathcal S$ is closed under tensor products where the monoidal structure on $Mod_\mathcal S(Pr^L)_{/\mathcal S}$ is the usual one on $C_{/A}$ where $A$ is an algebra: the tensor product is given by $(X\to A)\otimes (Y\to A) := (X\otimes Y\to A\otimes A\to A)$. You can make this more precise and actually make this into a monoidal category (as monoidal as $A$ is in $C$). It's closed under tensor products, but also every functor there is both in $Pr^R$ and in $Pr^L$.
3- The structure you're interested in is the structure of an $E_n$-algebra on $Mod(A)\to \mathcal S$, viewed in $Mod_\mathcal S(Pr^L)_{/\mathcal S}$. By the point just above, this is the same as the structure of an $E_n$-coalgebra in $Mod_\mathcal S(Pr^L)_{\mathcal S/}$ on the object $\mathcal S\overset{-\otimes A}\to Mod(A)$
4- Lurie proves in Higher Algebra that $Alg(\mathcal S)\to Mod_\mathcal S(Pr^L)_{\mathcal S/}$ is an equivalence of symmetric monoidal $\infty$-categories. So the structure you're after is exactly that of an $E_n$-coalgebra in algebras.
For $n=1$ you recover the notion of bialgebra. For $n=2$, I'm not sure what a quasitriangular bialgebra is, but if your condition is an "iff", then it should be the same data as an algebra $A$ together with a comultiplication $A\to A\otimes A$ (which is an algebra map) which is $E_2$-cocommutative as an algebra map, in a highly coherent sense. What comes later is just more and more cocommutativity for this map.
