Independence is sufficient but not necessary for the generating function of the sum of two random variables to be the product of their individual generating functions.
I am trying to come up with an example of two dependent rv's whose joint generating function is the product of the individual generating functions.
I have tried U:=X+Y and V:=X-Y where U and V are dependent but X and Y aren't. I ended up with $[\phi(t)]^2$ vs $\phi(t^2)$
Generating function $\phi_{X}(t) := \sum_{k = 0}^\infty Pr\{X=k\} t^k$
This is actually a necessary and sufficient condition (in the case of the joint generating function). Let $X$ and $Y$ have joint generating function $$\phi_{X,Y}(s,t) = \sum_{i,j=0}^\infty \mathbb P(X=i,Y=j)s^i t^j.$$ The product of the marginal generating functions is $$ \begin{align*} \phi_X(s)\phi_Y(t) &= \left(\sum_{i=0}^\infty\mathbb P(X=i)s^i\right)\left(\sum_{j=0}^\infty\mathbb P(Y=j)t^j\right)\\ &= \sum_{i,j=0}^\infty \mathbb P(X=i)\mathbb P(Y=j)s^it^j. \end{align*} $$ If we assume these are equal, then $$\mathbb P(X=i,Y=j)=\mathbb P(X=i)\mathbb P(Y=j)$$ for all $i,j\geqslant 0$, so $X$ and $Y$ are independent.
However, it is not the case that $$ \phi_{X,Y}(s,s) = \phi_X(s)\phi_Y(s)$$ implies that $X$ and $Y$ are independent. This is a subtle distinction, but $\phi_{X,Y}(s,s)$ is the generating function of $X+Y$, not the joint generating function of $(X,Y)$.
For a counterexample, let $X$ and $Y$ each be uniformly distributed over $\{0,1,2\}$, with joint distribution
$$ \mathbb P(X=i,Y=j) = \begin{cases} \frac19,& (i,j)\in\{(0,0), (1,1), (2,2)\}\\ \frac29,& (i,j)\in\{(0,2), (1,0), (2,1)\}\\ 0,&\text{otherwise} \end{cases} $$ Then $$\phi_X(s)\phi_Y(s)=\frac13(1+s+s^2)\frac13(1+s+s^2) = \frac1{9}(1+s+s^2)^2$$ and $$ \begin{align*} \phi_{X,Y}(s,s) &= \frac19 + \frac29 s + \frac39 s^2 + \frac29 s^3 +\frac19 s^4\\ &= \frac19(1 + 2s + 3s^2 + 2s^3 + s^4)\\ &= \frac19(1+s+s^2)^2\\ &= \phi_X(s)\phi_Y(s). \end{align*} $$ But $X$ and $Y$ are not independent, as for example $$\mathbb P(X=0)\mathbb P(Y=1)=\frac19\ne0=\mathbb P(X=0,Y=1).$$