Stable distributions that can be multiplied? Stable distributions are invariant under convolutions. What sub-families $F$ of the stable distributions are also closed under multiplication? In the sense that if $f\in F$ and $g\in F $, then the product probability density function, $f \cdot g$ (up to a normalization constant) also belongs to $F$?
Note: I substantially changed the content of this question. But the idea is essentially the same, and now it is much simpler. I only had a partial answer, so I think it's okay.
 A: A "stable distribution" is a particular kind of location-scale family of distributions.  The class of stable distributions is parameterized by two real numbers, the stability $\alpha\in(0,2]$ and skewness $\beta\in[-1,1]$.
A result quoted in the Wikipedia article resolves this question about closure under products of density functions.  When $f$ is the density of a stable distribution with $\alpha \lt 2$, then asymptotically
$$f(x) \sim |x|^{-(1+\alpha)} g(\operatorname{sgn}(x), \alpha, \beta)$$
for an explicitly given function $g$ whose details do not matter.  (In particular, $g$ will be nonzero either for all positive $x$ or all negative $x$ or both.)  The product of any two such densities therefore will be asymptotically proportional to $|x|^{-2(1+\alpha)}$ in at least one tail.  Since $2(1+\alpha)\ne 1+\alpha$, this product (after renormalization) cannot correspond to any distribution in the same stable family.
(Indeed, because $3(1+\alpha) \ne 1+\alpha^\prime$ for any possible $\alpha^\prime\in(0,2]$, the product of any three such density functions cannot even be the density function of any stable distribution.  That destroys any hope of extending the idea of product closure from a single stable distribution to a set of stable distributions.)
The only remaining possibility is $\alpha=2$.  These are the Normal distributions, with densities proportional to $\exp(-(x-\mu)^2/(2\sigma^2))$ for the location and scale parameters $\mu$ and $\sigma$.  It is straightforward to check that a product of two such expressions is of the same form (because the sum of two quadratic forms in $x$ is another quadratic form in $x$).
The unique answer, then, is that the Normal distribution family is the only product-of-density-closed stable distribution.
A: I know this is a partial answer and I'm not an expert, but this might help: if one of two unimodal pdfs is log-concave, then their convolution is unimodal. Due to Ibragimov (1956), via these notes. Apparently, if both are log-concave, then the convolution is also log-concave.
As far as product closure, the only "clean" result I know of for product distributions is the limit theorem described in this math.se answer.
How about a truncated version of these? The bounded uniform distribution is a limiting case of its shape parameter, and as far as I'm aware they're unimodal and log-concave so they have unimodal, log-concave convolutions. I have no clue about their products . When I have more time later this week I could try and run some simulations to see if I get log-concave products of truncated error distributions. Maybe Govindarajulu (1966) would help.
I'm not sure what the policy on crossposting is, but it seems like the math.se people might be able to help you as well. Out of curiosity, are you trying to build an algebraic structure out of probability distributions?
