This is not true without the assumption that $G$ is reduced. Here is a counterexample.
Fix a prime number $p$ and any field $k$ of characteristic $p$. We define the nonreduced algebraic subgroup $\alpha_p \subset \mathbb{G}_a$ by $\alpha_p = Spec(k[t]/(t^p))$ (here $\mathbb{G}_a$ denotes the additive group over $k$). Then $\alpha_p$ contains a single $k$-point-- the identity $0$ of the group.
We set $X = \mathbb{G}_a$. The group $\alpha_p$ acts on $X$ by addition (include $\alpha_p$ into $\mathbb{G}_a$ and then use the group structure).
The identity morphism $id: X \to X$ is not $\alpha_p$-invariant (the scheme-theoretic image of $\alpha_p \times 0 \subset \alpha_p \times X$ is just $0\subset X$ under the projection and $\alpha_p \subset X$ under the action). However, it is clearly invariant under the unique $k$-point of $\alpha_p$ (the identity of the group).
If $G$ is reduced and $S$ is separated, then I think that you should be fine (under your assumption that $k$ is algebraically closed).
Edited:
For the last part, you can argue as follows. Consider the morphism $$ \bigsqcup_{p \in G(k)} p \to G$$ By the Nullstellensatz, this has schematic image $G$. Using an argument similar to the end of the proof of Prop 3.2.4 (ii) in http://virtualmath1.stanford.edu/~conrad/249BW16Page/handouts/alggroups.pdf, you can show by arguing affine locally that the product morphism $$\bigsqcup_{p \in G(k)} p \times X \to G \times X$$ has schematic image $G \times X$. (Notice that the original morphism is not quasicompact! so it is not automatic that the schematic image commutes with flat base-change, but in this case it does. This is true when your base is a field because any ring over $k$ is a free $k$-module).
Now, if $S$ is separated, the locus of $G \times X$ where the projection agrees with the action is a closed subscheme $Z \subset G \times X$. By assumption, we have a factorization $$\bigsqcup_{p \in G(k)} p \times X \to Z \to G \times X$$ Therefore, by the scheme theoretic density we must have $Z = G \times X$.