= Solution
The <probability generating function> is
$$
p_X(z)=\mathbb E[z^X]=\sum_{n\geq1}\mathbb P(X=n)z^n,\qquad |z|\leq1.
$$
For <independent random variables>, the bounded factors $z^X$ and $z^Y$ have factorizing <expectation>, so
$$
\boxed{p_{X+Y}(z)=\mathbb E[z^Xz^Y]=\mathbb E[z^X]\mathbb E[z^Y]=p_X(z)p_Y(z).}
$$
This includes complex $z$ in the indicated disk by absolute convergence.
For the dice construction, encode the labels of an ordinary die by $F(z)=z+z^2+\cdots+z^6$. Its <probability generating function> is $F(z)/6$. A useful factorization into <polynomial factors> is
$$
F(z)=z(1+z)(1+z+z^2)(1-z+z^2).
$$
Use <redistribution of polynomial factors for dice> to define
$$
\begin{aligned}
A(z)&=\frac{F(z)}{1-z+z^2}=z(1+z)(1+z+z^2)=z+2z^2+2z^3+z^4,\\
B(z)&=F(z)(1-z+z^2)=z+z^3+z^4+z^5+z^6+z^8.
\end{aligned}
$$
Both <polynomials> have nonnegative integer <coefficients>, no constant term, and coefficient sum six. They therefore describe fair six-sided dice with strictly positive face labels
$$
\boxed{A:(1,2,2,3,3,4),\qquad B:(1,3,4,5,6,8).}
$$
When these dice are thrown with <independent> outcomes, the sum has <probability generating function> $A(z)B(z)/36=F(z)^2/36$. Equality of <coefficients> proves equality of every sum <probability>, including all totals from two to twelve, with zero probability outside that range. Both dice are nonstandard, completing the construction.
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