2 corrected major errors

Edit: Somehow I totally misread the question. I talked about the group algebra $\mathbb C[G]$, which is not at all the same as the character ring $R(G)$. Over $\mathbb C$ (or any other field of characteristic 0), $R(G)$ is naturally a subalgebra of $(\mathbb C[G])^*$, which is the algebra of functions on $G$ with pointwise multiplication, and now the comultiplication encodes the group structure. On the other hand, it is not a subbialgebra: the coproduct of a class function is not a class function.

Anyway, original post below, with the obviously wrong things struck out. So it's really an answer to Kevin, rather than anything else.

Well, it depends on what you mean by "$R(G)$". I won't address TK duality, and most of what I'll say is essentially a follow-up to Kevin's answer, rather than an answer in its own right. Also, I'm only going to address finite groups and their finite-dimensional representations. Also, for me the word "ring" means (associative, unital, noncommutative) "$\mathbb C$-algebra".

Recall that a complex representation of $G$ is the same as an algebra representation of $\mathbb C[G]$. Let $R$ be a ring. As Kevin says, it's in general impossible to define an $R$-module structure on $M\otimes N$ when $M,N$ are $R$-modules. (When $R$ is abelian, which is not the case here, one can define a tensor product $M \otimes_R N$, but that's not the tensor product of representations anyway.) What would a tensor product of modules require? It would require a rule that assigns to each $r\in R$ and each pair $M,N$ of $R$-modules an endomorphism of $M\otimes N$, of course, and we should impose all sorts of axioms that force the tensor product to be well-behaved. Among other things, it's much easier if the endomorphism is an element of the tensor product $\text{End}(M) \otimes \text{End}(N) \subseteq \text{End}(M\otimes N)$. And we already have some distinguished elements of $\text{End}(M)$ and $\text{End}(N)$, namely the action of $R$.

So one way to try to construct a well-behaved tensor product on the category of $R$-modules is to find a nice map $\Delta: R \to R\otimes R$. Then the axioms for this map that assure that the tensor product is good are that $\Delta$ be an algebra homomorphism, and that it be "coassociative": $(\text{id}\otimes \Delta)\circ \Delta = (\Delta \otimes \text{id})\circ \Delta)$. Let's suppose that there's also a distinguished "trivial" representation $\epsilon: R \to \text{End}(\mathbb C) = \mathbb C$; if this is to be the monoidal unit, then we'd need $(\text{id}\otimes \epsilon) \circ \Delta = \text{id} = (\epsilon \otimes \text{id})\circ \Delta$. The maps $\Delta, \epsilon$ satisfying these axioms define on $R$ the structure of a bialgebra.

By the way, the map is called "$\Delta$" because if $G$ is a group (or monoid) and $R = \mathbb C[G]$, then the map $R \to R\otimes R$ given on the basis $G$ by the diagonal map $\Delta: g \mapsto g\otimes g$ is such a structure.

Then here's a cool fact. Define an element $r\in R$ to be grouplike if $\Delta(r) = r\otimes r$. Then the grouplike elements are a multiplicative submonoid of $R$. And when $R = C[G]$, the grouplike elements are precisely $G$.

So my answer to your question is that "the additional information contained in $R(G)$ as opposed to the character table" is its bialgebra structure. As the other interlocutors have pointed out, the ring structurehas no extra information.

1

Well, it depends on what you mean by "$R(G)$". I won't address TK duality, and most of what I'll say is essentially a follow-up to Kevin's answer, rather than an answer in its own right. Also, I'm only going to address finite groups and their finite-dimensional representations. Also, for me the word "ring" means (associative, unital, noncommutative) "$\mathbb C$-algebra".

Recall that a complex representation of $G$ is the same as an algebra representation of $\mathbb C[G]$. Let $R$ be a ring. As Kevin says, it's in general impossible to define an $R$-module structure on $M\otimes N$ when $M,N$ are $R$-modules. (When $R$ is abelian, which is not the case here, one can define a tensor product $M \otimes_R N$, but that's not the tensor product of representations anyway.) What would a tensor product of modules require? It would require a rule that assigns to each $r\in R$ and each pair $M,N$ of $R$-modules an endomorphism of $M\otimes N$, of course, and we should impose all sorts of axioms that force the tensor product to be well-behaved. Among other things, it's much easier if the endomorphism is an element of the tensor product $\text{End}(M) \otimes \text{End}(N) \subseteq \text{End}(M\otimes N)$. And we already have some distinguished elements of $\text{End}(M)$ and $\text{End}(N)$, namely the action of $R$.

So one way to try to construct a well-behaved tensor product on the category of $R$-modules is to find a nice map $\Delta: R \to R\otimes R$. Then the axioms for this map that assure that the tensor product is good are that $\Delta$ be an algebra homomorphism, and that it be "coassociative": $(\text{id}\otimes \Delta)\circ \Delta = (\Delta \otimes \text{id})\circ \Delta)$. Let's suppose that there's also a distinguished "trivial" representation $\epsilon: R \to \text{End}(\mathbb C) = \mathbb C$; if this is to be the monoidal unit, then we'd need $(\text{id}\otimes \epsilon) \circ \Delta = \text{id} = (\epsilon \otimes \text{id})\circ \Delta$. The maps $\Delta, \epsilon$ satisfying these axioms define on $R$ the structure of a bialgebra.

By the way, the map is called "$\Delta$" because if $G$ is a group (or monoid) and $R = \mathbb C[G]$, then the map $R \to R\otimes R$ given on the basis $G$ by the diagonal map $\Delta: g \mapsto g\otimes g$ is such a structure.

Then here's a cool fact. Define an element $r\in R$ to be grouplike if $\Delta(r) = r\otimes r$. Then the grouplike elements are a multiplicative submonoid of $R$. And when $R = C[G]$, the grouplike elements are precisely $G$.

So my answer to your question is that "the additional information contained in $R(G)$ as opposed to the character table" is its bialgebra structure. As the other interlocutors have pointed out, the ring structure has no extra information.