I think the answer to the original question (i.e. are there any universal algebraic identities relating convolution and multiplication over arbitrary groups, beyond the "obvious" ones?) is negative, though establishing it rigorously is going to be tremendously tedious.
There are a couple steps involved. To avoid technicalities let's restrict attention to discrete finite fields G (so we can use linear algebra), and assume the characteristic of G is very large.
Firstly, given any purported convolution/multiplication identity relating a bunch of functions, one can use homogeneity and decompose that identity into homogeneous identities, in which each function appears the same number of times in each term. (For instance, if one has an identity involving both cubic expressions of a function f and quadratic expressions of f, one can separate into a cubic identity and a quadratic identity.) So without loss of generality one can restrict attention to homogeneous identities.
Next, by depolarisation, one should be able to reduce further to the case of multilinear identities: identities involving a bunch of functions $f_1, f_2, ..., f_n$, with each term being linear in each of the functions. (I haven't checked this carefully but it should be true, especially since we can permit the functions to be complex valued.)
It is convenient just to consider evaluation of these identities at the single point 0 (i.e. scalar identities rather than functional identities). One can actually view functional identities as scalar identities after convolving (or taking inner products of) the functional identity with one additional test function.
Now (after using the distributive law as many times as necessary), each term in the multilinear identity consists of some sequence of applications of the pointwise product and convolution operations (no addition or subtraction), evaluated at zero, and then multiplied by a scalar constant. When one expands all of that, what one gets is a sum (in the discrete case) of the tensor product $f_1 \otimes ... \otimes f_n$ of all the functions over some subspace of $G^n$. The exact subspace is given by the precise manner in which the pointwise product and convolution operators are applied.
The only way a universal identity can hold, then, is if the weighted sum of the indicator functions of these subspaces (counting multiplicity) vanishes. (Note that finite linear combinations of tensor products span the space of all functions on $G^n$ when $G$ is finite.)
But when the characteristic of $G$ is large enough, the only way that can happen is if each subspace appears in the identity with a net weight of zero. (Indeed, look at a subspace of maximal dimension in the identity; for $G$ large enough characteristic, it contains points that will not be covered by any other subspace in the identity, and so the only way the identity can hold is if the net weight of that subspace is zero. Now remove all terms involving this subspace and iterate.)
So the final thing to do is to show that a given subspace can arise in two different ways from multiplication and convolution only in the "obvious" manner, i.e. by exploiting associativity of pointwise multiplication and of convolution. This looks doable by an induction argument but I haven't tried to push it through.
Best Answer
If and only if the Picard group is trivial. (Similar to in algebraic number theory).
Proof: Suppose $R$ is a UFD. Then every codimension 1 prime ideal is principal. Since these generate the divisor group, every divisor is principal. Suppose $R$ is not a UFD. Take an irreducible element that is not prime. This element must be contained in a codimension 1 prime ideal that is not principal (I think?). This would be a non-principal divisor.
There may be some assumptions I'm forgetting on the divisor group.
If you think the Picard group is geometric enough, we are done. (In particular, there are topological obstructions to having trivial Picard group.)
EDIT: Francois Brunault points out that these arguments are about the Weil divisor class group, not the Picard group. These concepts agree on nonsingular varieties but differ on singular ones. Sadly, the Weil group is less geometric than the Picard group.