The very short answer is yes, provided that you also allow yourself a little linear algebra. But then again you rejected David's answer, so you may not be happy with mine. I'll try to convince you that my answer is both trivial and also deep, and that it doesn't depend on more structure than what you've allowed.
The short answer
For the purposes of my answer, I will pretend that the group $G$ is finite (I won't pretend it's abelian, because I don't need it to be). There are versions of what I'm going to say for, at least, compact groups and algebraic groups, but subtleties emerge which I will ignore. Let $R$ be the ring of functions on $G$. Since $G$ is finite, $R$ is finite-dimensional. (If $G$ is algebraic, $R$ is like a polynomial ring, and if $G$ is compact, $R$ has a good topology, and any constructions must be completed. This is what I mean by "subtleties".)
The ring of functions on $G$ has a canonical nondegenerate pairing: $\langle f,g\rangle = \int fg$. Being nondegenerate, the pairing has an "inverse", which is an element of the tensor product $R\otimes R$. Explicitly, pick any orthonormal basis of the pairing, e.g. the basis $\{\delta_x: x\in G\}$, where $\delta_x(y)=1$ if $y=x$ and $0$ otherwise. Then the inverse is the sum over the basis of the tensor square of each item. So for my basis, it is $\sum_{x\in G} \delta_x\otimes\delta_x$. But it should be emphasized that the inverse to the pairing does not actually depend on the basis. Since I don't have better notation, though, I'll work in the basis for my answer. The better description is in terms of the physicists' abstract index notation, or Penrose's birdtracks.
Then convolution and multiplication are related by the following:
$$\langle f_1\cdot f_2, f_3*f_4\rangle = \sum_{x_1} \sum_{x_2} \sum_{x_3} \sum_{x_4} \langle f_1, \delta_{x_1}*\delta_{x_2}\rangle \langle f_2, \delta_{x_3}*\delta_{x_4}\rangle \langle \delta_{x_4}\cdot\delta_{x_2}, f_3\rangle \langle \delta_{x_3}\cdot\delta_{x_1}, f_4\rangle $$
The long answer
Let $X$ be a set (or more generally a "space"). Write $C(X)$ for the ring of functions on $X$, and $\mathbf k[X]$ for the collection of linear combinations of points in $X$. (I'll write $\mathbf k$ for the ground field; everything I say will work over any field, but you can take it to be the reals or complexes if you want. The meaning of the word "space" probably depends on your ground field.)
What types of operations are these? Well, $C()$ is a contravariant functor, and $\mathbf k[]$ is a covariant one, both from the category of SETS (or SPACES) to the category VECT. Let's start with $\mathbf k[]$ because it's covariant. It's actually better than a functor: it's a monoidal functor, meaning that if you start with a cartesian product you end up with a tensor product: $k[X\times Y] = \mathbf k[X\otimes Y]$. Actually, so is $C()$, although if you work with non-finite sets you have to complete the tensor product. In fact, these two operations are intimately related: for any space $X$, $C(X)$ is naturally (i.e. functorially in $X$) the dual space to $\mathbf k[X]$, so that $C() = \mathbf k[]^*$. Thus, we can basically completely understand $C()$ by understanding $\mathbf k[]$, or vice versa.
Since SETS (or SPACES) is cartesian, every object is a coalgebra in a unique way. I'll spell this out. An algebra in a monoidal category is an object $X$ with a "multiplication" map $X \times X \to X$ satisfying various conditions. The word is chosen so that in VECT, my notion of algebra and yours match. In SET, an algebra is a monoid. Anyway, a coalgebra is whatever you get by turning all the arrows around. For intuition, think about VECT, where a coalgebra is whatever the natural structure on the dual vector space to an algebra is. (Write the multiplication map as a big matrix from the tensor square of your algebra to your algebra, and think about its transpose.)
The canonical coalgebra structure on a set $X$, by the way, is given by the diagonal map $\Delta : X \to X \times X$, where $\Delta(x) = (x,x)$.
Since $\mathbf k[]$ is a monoidal functor, it takes algebras to algebras and coalgebras to coalgebras. Thus for any set $X$, the vector space $\mathbf k[X]$ inherits a coalgebra structure. Thus, dually, $C(X)$ inherits an algebra structure (you can say this directly: a monoidal contravariant functor turns coalgebras into algebras). In fact, this is precisely the canonical algebra structure you're calling "." on the ring of functions.
Well, let's say now that $X$ is an algebra in SETS, i.e. a monoid (e.g. a group). Then $\mathbf k[X]$ inherits an algebra structure, and equally $C(X)$ has a coalgebra structure. But actually it's a bit better than this. Since any set is a coalgebra in a unique way, the algebra and coalgebra structures on $X$ get along. I'll write $*$ for the multiplication in $X$. Then when I say "get along" what I mean is:
$$\Delta(x) * \Delta(y) = \Delta(x*y)$$
where on the left-hand-side I mean the component-wise multiplication in $X \times X$.
Well, $\mathbf k[]$ is a functor, so it preserves this equation, except that the coalgebra structure on $\mathbf k[X]$ is not trivial the way $\Delta$ is in SETS. Anything that is both a coalgebra and an algebra and that satisfies an equation like the one above is a bialgebra. You can check that the equation is well-behaved under dualizing, so that $C(X)$ is also a bialgebra if $X$ is an algebra.
Ok, so how does all this connect with your question? What's going on is that for sufficiently good spaces, e.g. finite sets, there is a canonical identification between the vector spaces $\mathbf k[X]$ and $C(X)$ for any $X$. This identification breaks various functoriality properties, though. But anyway, if $G$ is a finite group, then we can consider $\mathbf k[G]$ and $C(G)$ to be the same vector space $R$, and pretend that it just has two separate ring structures on it.
But doing this obscures the bialgebra property. If I'm only allowed to reference the two multiplications, and not their dual maps, then to write the bialgebra property requires explicitly referring to the canonical pairing (what I called $\int = \langle,\rangle$ before) and its inverse. Then the bialgebra property becomes the long equation I wrote in the previous part.
Final remarks
I should also mention that a group has not just a multiplication but also identities and inverses. These give another equation. In the basis from the first section, the unit in $R$ for $\cdot$ is the function $1 = \sum_{x\in G} \delta_x$, and the unit for $*$ is $\delta_e$, where $e$ is the identity in $G$. These satisfy the equation:
$$\delta_e \otimes 1 = \sum_{x_1} \sum_{x_2} (\delta_{x_1} * \delta_{x_2}) \otimes (\delta_{x_1} \cdot \delta_{x_2^{-1}})$$
where ${x_2^{-1}}$ is the inverse element to $x_2$. You should be able to recognize the inverse to the canonical pairing in there. Again, the equation is simpler in better notation, e.g. indices or birdtracks, and does not depend on a choice of basis. A bialgebra satisfying an equation like the one above is a Hopf algebra.
Another thing I should mention is that there are similar stories at least for compact groups, but you have to think harder about what "the inverse to the canonical pairing" is. (On a compact group, there is a canonical pairing of functions, given by Haar measure.) In fact, I think a story like this can be told for other spaces, where you change what you mean by $\mathbf k[]$ and $C()$, in the first case expanding the notion of linear combination and in the second case restricting the type of function. Then you should put the word "quasi" in front of everything, because the coalgebra structure, the inverse to the pairing, the units, etc. all require completions of your vector spaces.
And there may be special equations for abelian groups. In abelian land, the Fourier/Pontryagin transform does the following: it recognizes the (now commutative) ring $\mathbf k[G]$ as a ring of functions on some other space: $\mathbf k[G] = C(G^*)$.
But the overall moral is that really convolution and multiplication are going on in different vector spaces; it's just that you have a canonical pairing that you can't tell the spaces apart. And if you insist on conflating the two spaces, then you should allow the canonical pairing and its inverse as basic algebraic operations.
