**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.

Character Theory of Finite Groups, exercises 2.4 and 2.5, there are also some differences between $D_4$ and $Q_8$. $\endgroup$ – user19475 Jan 27 '18 at 13:37