Just multiply the two $D_6(x)$ generating functions and refactor them:
$$D_6(x)D6(x) = x^2 (1 + x)^2 (1 - x + x^2)^2 (1 + x + x^2)^2$$
Then multiply groups of factors so you get two polynomials with 6 terms, no negative no constant terms:
$$D_6(x)D_6(x)= (x+x^2+x^2+x^3+x^3+x^4)(x+x^3+x^4+x^5+x^6+x^8)$$
$$\Rightarrow \{1,2,2,3,3,4\}, \{1,3,4,5,6,8\}$$
While there may be some cute trick to this, I used (and would recommend) Mathematica or a similar CAS to expand and refactor polynomials.
The Mathematica process, by request (the process should work in any CAS, probably even W|A).
$$\textrm{factor}\left[(x+x^2+x^3+x^4+x^5+x^6)^2\right] = x^2(x+1)^2(x^2-x+1)^2(x^2+x+1)^2$$
What you want to do is multiply the factors to make $D_6^*(x), D_6^{**}(x)$ so that you have a six term polynomial without negative or constant terms.
The makesure that it has six terms (6 faces) we want $D_6^*(1) = D_6^{**}(1) = 6$. So consider the following at $x=1$:
$$x \mapsto 1$$
$$(x+1) \mapsto 2$$
$$(x^2-x+1) \mapsto 1$$
$$(x^2+x+1) \mapsto 3$$
So we need to give each of the polynomials one $(x^2+x+1)$ and one $(x+1)$ term so the product will be six when evaluated at one.
To remove the constant term, split the $x^2$ between the two polynomials.
These leave us with the following:
$$D_6^*(x) = x(x+1)(x^2+x+1)j(x), D_6^{**}(x) = x(x+1)(x^2+x+1)k(x)$$
Where $k(x)$ and $j(x)$ are some polynomials that we can create from the remaining factors. However, we have two copies of $(x^2-x+1)$ remaining, and if we set $k(x) = j(x)$, then we will simply have the original dice back, so both of them go to one of the nonstandard dice. WLOG, set $k(x) = 1$ and $j(x) = (x^2-x+1)^2$.
Our final result:
$$D_6^*(x) = x(x+1)(x^2+x+1)(x^2-x+1)^2, D_6^{**}(x) = x(x+1)(x^2+x+1)$$
Then in Mathematic:
$$\textrm{expand}\left[x(x+1)(x^2+x+1)(x^2-x+1)^2\right] = x+x^3+x^4+x^5+x^6+x^8$$
$$\textrm{expand}\left[x(x+1)(x^2+x+1)\right] = x+x^2+x^2+x^3+x^3+x^4$$
As before, the exponents are the sides of each die, so we just read those off.
Full disclosure: I didn't go through much of a process beyond distributing the $x$ to both of them the first time; the GF is short enough that I came up with that answer after trying a few combinations. For longer GFs, you'll need to do a similar analysis to add enough contraints to make it feasible to solve.
