How about if $A = E_1\times E_2$ with $E_i$ non isotrivial elliptic curves and $s: S \to P\times 0$. Then your $\mathbb Z.X$ is contained in $E_1\times 0$ and so is not zariski dense. In fact, this is the only thing that can happen.

Let $Z$ be the Zariski closure of $\mathbb Z.X$ in $A$. Then $Z$ is closed under addition and inversion and therefore the connected component of $Z$ is a sub abelian variety. If $X$ is non-torsion, then this connected component will be of positive dimension. Conversely, if we had such a sub abelian variety, any section contained in this subvariety will give rise to a non zariski dense $\mathbb Z. X$.