All right, I think I can finish the proof now. I will prove that there are no nontrivial identities on $\mathbb{R}^d$ for any $d$. This proof makes heavy use of the first part of Terry Tao's post (reducing to multilinear identities), but I'll use a different argument to finish it, since I guess I'm just more familiar and comfortable with real vector spaces than with finite groups. It should be possible to complete Terry's line of argument to get a proof for sufficiently large finite groups, which my proof won't cover. Moreover, as Theo pointed out in a comment to his answer, deforming the domain nonlinearly screws up convolution while leaving the other operations intact, and it should be easy to use that to show no identities can hold. In any case, this is a community wiki post, so anybody can make additions or simplifications.

First, by Terry Tao's observations, it suffices to consider multilinear identities of the form
$$c_1F_1(f_1,\ldots,f_n) + \cdots + c_kF_k(f_1,\ldots,f_n) = 0$$ where each $F_i$ is a
"multilinear monomial," i. e., a composition of multiplication and convolution in which each of $f_1,\ldots,f_n$ appears exactly once. (The original question didn't allow scalar multiplication, but it doesn't introduce any difficulty.) To summarize the argument: by applying the distributive laws as much as necessary and using an easy scaling argument, it suffices to consider identities that are homogeneous in each argument, i. e., sums of monomials in which each argument appears some fixed number of times. To reduce this further to the multilinear case, suppose we have some putative identity of the form $F(f_1,\ldots,f_m) = 0$ that is homogenous of degree $n_i$ in $f_i$ for all $i$. For the moment, consider $f_2,\ldots,f_n$ to be fixed, so we have a homogeneous degree-$n_1$ functional $T(f_1)$ of $f_1$. The polarization identity states that if we define a new functional $S$ by
$$S(g_1,\ldots,g_{n_1}) = \frac{1}{n_1!}\sum_{E\subseteq \{1,\ldots,n\_1\}} (-1)^{n_1-|E|}
T\big(\sum_{j\in E} g_j\big),$$ then $S$ is a (symmetric) multilinear function of $g_1,\ldots,g\_{n_1}$ and $S(f_1,\ldots,f_1) = T(f_1)$. Thus, the identity $F(f_1,\ldots,f_m)=0$ is equivalent to the identity $G(g_1,\ldots,g_{n_1},f_2,\ldots,f_n) = 0$, where $G$ is obtained from $F$ by the polarization construction applied on the first argument. Repeating the construction for $f_2,\ldots,f_m$, we obtain an equivalent multilinear identity $H(g_1,\ldots,g_n)=0$ (where $n=n_1+\cdots+n_m$).

Let's fix a nomenclature for monomials: let $C(f_1,\ldots,f_n)=f_1*\cdots*f_n$ and $M(f_1,\ldots,f_n)=f_1\cdot \cdots \cdot f_n$. A monomial is a **C-expression** if convolution is the top-level operation or an **M-expression** if multiplication is the top-level expression. $f_1,\ldots,f_n$ are **atomic expressions** and are considered both M-expressions and C-expressions. We consider two monomials to be identical if they can be obtained from one another by applying the associative and commutative laws for multiplication and convolution. With this equivalence relation, each equivalence class of monomials can be written uniquely in the form $C(A_1,\ldots,A_n)$ or $M(B_1,\ldots,B_n)$ (up to permuting the $A$s or the $B$s), where the $A$s are M-expressions and the $B$s are C-expressions. At this point, we have made maximal use of the algebraic identities for the convolution algebra and the multiplication algebra, so now we have to prove that there are no identities whatsoever of the form
$$c_1F_1(f_1,\ldots,f_n) + \cdots + c_kF_k(f_1,\ldots,f_n) = 0$$
where the $c_i$ are nonzero scalars and the $F_i$ are distinct multilinear monomials.

For all $a>0$, let $\phi_a:\mathbb{R}^d\to \mathbb{R}$ be the gaussian function $\phi_a(x)=e^{-a\|x\|^2}$. We'll prove if the $F_i$ are distinct and the $c_i$ are nonzero, then
$$c_1F_1(\phi_{a_1},\ldots,\phi_{a_n}) + \cdots + c_kF_k(\phi_{a_1},\ldots,\phi_{a_n})= 0$$ cannot hold for all $a_1,\ldots,a_n>0$.
It's easy to see that $\phi_a\cdot\phi_b = \phi_{a+b}$ and $\phi_a*\phi_b = (\pi(a+b))^{d/2}\phi_{(a^{-1}+b^{-1})^{-1}}$. Therefore, if we define $S(a_1,\ldots,a_n)=a_1+\cdots +a_n$ and $P(a_1,\ldots,a_n)=(a_1^{-1}+\cdots+a_n^{-1})^{-1}$, and $F$ is a multilinear monomial, then $F(\phi_{a_1},\ldots,\phi_{a_n}) = R_F(a_1,\ldots,a_n)^{d/2}\exp(-Q_F(a_1,\ldots,a_n)\|x\|^2)$, where $R_F$ is a rational function and $Q_F$ is a rational function composed of $S$ and $P$. In fact, if $F$ is written as a composition of $C$ and $M$, then $Q_F(a_1,\ldots,a_n)$ is obtained from $F(\phi_{a_1},\ldots,\phi_{a_n})$ simply by replacing all the $C$s by $P$s, the $M$s by $S$s, and $\phi_{a_i}$ by $a_i$ for all $i$. Therefore, it makes sense to define P- and S-expressions analogously to C- and M-expressions. A **PS-expression** in $a_1,\ldots,a_n$ is a composition of $P$ and $S$ in which each of $a_1,\ldots,a_n$ appears exactly once. Equivalence of PS-expressions is defined exactly as for C/M monomials; in particular, equivalence of PS-expressions is apparently a stronger condition than equality as rational functions.

The main lemma we need is that it actually isn't a stronger condition: if $F$ and $G$ are distinct multilinear monomials in $n$ arguments, then $Q_F$ and $Q_G$ are distinct rational functions. In other words, distinct PS-expressions define distinct rational functions. (Note that this is false if the adjective "multilinear" is dropped.) To prove this, first note that although $Q_F$ and $Q_G$ are initially defined as functions $(0,\infty)^n\to (0,\infty)$, they extend continuously $[0,\infty)^n\to [0,\infty)$. If $D=\{i_1,\ldots,i_k\}$ is a subset of $\{1,\ldots,n\}$ and $Q$ is a PS-expression in $n$ variables, then $D$ is called a **prime implicant** of $Q$ if $Q(a_1,\ldots,a_n) = 0 $ when $a_{i_1},\ldots,a_{i_k}$ are all set to zero, but no proper subset of $D$ has this property. Let $I(Q)$ be the set of prime implicants of $Q$. It's easy to show that $I(P(Q_1,\ldots,Q_m))$ is the disjoint union of $I(Q_1),\ldots,I(Q_m)$, and $I(S(Q_1,\ldots,Q_m))$ is the set of all $D_1 \cup \cdots \cup D_k$, where $D_i\in I(Q_i)$. (It's important here that none of the variables $a_1,\ldots,a_n$ appears in more than one $Q_i$.) Define the **implicant graph** of $Q$ as the undirected graph with vertices $1,\ldots,n$ and an edge between $i$ and $j$ if some prime implicant of $Q$ contains both $i$ and $j$. It's easy to see that the implicant graph of an S-expression is connected, and if $Q_1,\ldots,Q_m$ are S-expressions, then the connected components of the implicant graph of $P(Q_1,\ldots,Q_m)$ are the implicant graphs of $Q_1,\ldots,Q_m$. This immediately implies that a P-expression cannot define the same function as an S-expression, so it suffices to show that distinct S-expressions induce distinct rational functions, and distinct P-expressions do. Actually, it suffices to show that distinct P-expressions define distinct expressions, since $P$ and $S$ are exchanged by the involution $\sigma(a)=a^{-1}$: $\sigma(P(a,b))=S(\sigma(a),\sigma(b))$. That different P-expressions induce different functions now follows by induction on the number of variables, since the implicant sets of the S-expressions $Q_i$ are uniquely determined by the implicant set of $P(Q_1,\ldots,Q_m)$ by considering connectivity as above.

The rest of the proof is easy: if the $F_i$ are distinct multilinear monomials, then the $Q_{F_i}$ are distinct rational functions. This implies that for some $a_1,\ldots,a_n$, the $Q_{F_i}(a_1,\ldots,a_n)$ are all distinct positive numbers, since distinct rational functions can't agree on a set of positive Lebesgue measure. To get a contradiction, suppose the $c\_i$ are all nonzero and the identity $\sum_i c_i F_i(f_1,\ldots,f_n)=0$ holds. Then
$$\sum_i c_i F_i(\phi_{a_1},\ldots,\phi_{a_n}) = \sum_i c_i R_{F_i}(a_1,\ldots,a_n)^{d/2} \exp(-Q_{F_i}(a_1,\ldots,a_n)\|x\|^2) = 0$$ for all $x$. Without loss of generality, the $Q_{F_i}(a_1,\ldots,a_n)$ are increasing as a function of $i$. But then for large enough $x$, the first term dominates all the others, so the sum can't be zero unless $c_1=0$: a contradiction. This completes the proof.

`f*(g.h) = (h.g)*f`

rather then`f*(g.h) = (f.g)*h`

(the last one is incorrect, of course). – Ilya Nikokoshev Oct 30 '09 at 19:18