The obstruction to $\mathrm{Pic}(S)$ being generated by invariant divisors is pretty much the Picard group of the generic $G$-orbit which, in turn, is controlled by the character group of the generic isotropy group. So for $S=\mathbf{P}^2$ acted on transitively by $G=GL(2)$ or $G=SL(2)$ the Picard group is certainly not generated by invariant divisors since there aren't any.
On the positive side let me tackle Q3 which might be a blueprint for for general cases. Let $G$ be a torus (not necessarily of dimension $1$) acting effectively on a normal variety $S$ (not necessarily of dimension $2$). Then I claim that $A^1(S)$ is generated by invariant divisors. To see this, observe that $S$ contains an invariant open subset $S_0$ of the form $Y\times T$ where $Y=S_0/T$ (use Rosenlicht for the existence of $Y$ and Hilbert Satz 90 for the triviality of the bundle $S_0\to Y$). Then $A^1(S_0)=A^1(Y)$ is generated by invariant divisors of the form $D\times T$. Thus $A^1(S)$ is genrated by their closures plus the irreducible components of $S\setminus S_0$ which are of codimension $1$ in $S$. The latter are $G$-invariant, as well.
