I couldn't make the above answer work, so here's an approach explained to me by Rune Haugseng. Let $C$ be symmetric monoidal and $p\colon C^\otimes\to Fin_\ast$ be the cocartesian fibration witnessing this. First notice that $CoAlg(C)^{op}\simeq Alg(C^{op})$ has a "pointwise" symmetric monoidal structure which is given by HA.3.2.4.3, and therefore $CoAlg(C)$ also has a symmetric monoidal structure i.e. there is a cocartesian fibration $q\colon CoAlg(C)^\otimes\to Fin_\ast$. Here I'm using the fact that $C^{op}$ has a symmetric monoidal structure induced by taking the fiberwise dual of $p$ in the sense of this paper. Similar considerations give a symmetric monoidal structure to $LCoMod(C)\simeq LMod(C^{op})^{op}$. Moreover, there is a cartesian fibration $LMod(C^{op})\to Alg(C^{op})$ by HA.4.2.3.2 which gives a fiberwise cartesian fibration $LMod(C^{op})^\otimes\to Alg(C^{op})^\otimes$. By taking fiberwise duals again we get a fiberwise cocartesian fibration $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$. It's a little tricky, but one can check that this fiberwise cocartesian fibration satisfies the condition A.1.8 of this paper and is therefore a cocartesian fibration. On a fixed fiber this says that if I've got a map of coalgebras $A\to B$ (or a finite list of maps of coalgebras) then I get a left adjoint functor $LCoMod_A(C)\to LCoMod_B(C)$ which takes the coaction $M\to A\otimes M$ to $M\to A\otimes M\to B\otimes M$ (this is the "opposite" of restriction of scalars). But now we've lifted it up to $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$ so that it plays well with the monoidal structure, which we'll need.

Now suppose that $H$ is a bialgebra in $C$, i.e. $H$ is an algebra object in $CoAlg(C)\simeq Alg(C^{op})^{op}$ . In other words, $H$ is determined by a functor of $\infty$-operads $H^\otimes\colon Assoc^\otimes\to CoAlg(C)^\otimes$. Then we can pull back the cocartesian fibration $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$ along $H^\otimes$ to obtain a cocartesian fibration $LCoMod_H(C)^\otimes\to Assoc^\otimes$. There's a little bit of checking to do if you want to make sure that this pullback really is equivalent to $LCoMod_H(C)^n$ over each $\langle n\rangle\in Assoc^\otimes$, but it's not too bad.

I should add that you can fully analyze the cocartesian morphisms in $LCoMod_H(C)^\otimes$ and check that the $H$-coaction on the tensor product of two $H$-comodules $M$ and $N$ is indeed $M\otimes N\to M\otimes H\otimes N\otimes H\to M\otimes N\otimes H\otimes H\to M\otimes N\otimes H$.

I couldn't make the above answer work, so here's an approach explained to me by Rune Haugseng (of course any errors are entirely my own). Let $C$ be symmetric monoidal and $p\colon C^\otimes\to Fin_\ast$ be the cocartesian fibration witnessing this. First notice that $CoAlg(C)^{op}\simeq Alg(C^{op})$ has a "pointwise" symmetric monoidal structure which is given by HA.3.2.4.3, and therefore $CoAlg(C)$ also has a symmetric monoidal structure i.e. there is a cocartesian fibration $q\colon CoAlg(C)^\otimes\to Fin_\ast$. Here I'm using the fact that $C^{op}$ has a symmetric monoidal structure induced by taking the fiberwise dual of $p$ in the sense of this paper. Similar considerations give a symmetric monoidal structure to $LCoMod(C)\simeq LMod(C^{op})^{op}$. Moreover, there is a cartesian fibration $LMod(C^{op})\to Alg(C^{op})$ by HA.4.2.3.2 which gives a fiberwise cartesian fibration $LMod(C^{op})^\otimes\to Alg(C^{op})^\otimes$. By taking fiberwise duals again we get a fiberwise cocartesian fibration $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$. It's a little tricky, but one can check that this fiberwise cocartesian fibration satisfies the condition A.1.8 of this paper and is therefore a cocartesian fibration. On a fixed fiber this says that if I've got a map of coalgebras $A\to B$ (or a finite list of maps of coalgebras) then I get a left adjoint functor $LCoMod_A(C)\to LCoMod_B(C)$ which takes the coaction $M\to A\otimes M$ to $M\to A\otimes M\to B\otimes M$ (this is the "opposite" of restriction of scalars). But now we've lifted it up to $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$ so that it plays well with the monoidal structure, which we'll need.

Now suppose that $H$ is a bialgebra in $C$, i.e. $H$ is an algebra object in $CoAlg(C)\simeq Alg(C^{op})^{op}$ . In other words, $H$ is determined by a functor of $\infty$-operads $H^\otimes\colon Assoc^\otimes\to CoAlg(C)^\otimes$. Then we can pull back the cocartesian fibration $LCoMod(C)^\otimes\to CoAlg(C)^\otimes$ along $H^\otimes$ to obtain a cocartesian fibration $LCoMod_H(C)^\otimes\to Assoc^\otimes$. There's a little bit of checking to do if you want to make sure that this pullback really is equivalent to $LCoMod_H(C)^n$ over each $\langle n\rangle\in Assoc^\otimes$, but it's not too bad.

I should add that you can fully analyze the cocartesian morphisms in $LCoMod_H(C)^\otimes$ and check that the $H$-coaction on the tensor product of two $H$-comodules $M$ and $N$ is indeed $M\otimes N\to M\otimes H\otimes N\otimes H\to M\otimes N\otimes H\otimes H\to M\otimes N\otimes H$.