Yes, such a partial order can be extended to a total order: we can amend the proof of the Szpilrajn extension theorem which essentially establishes the result in the case $G=1$.
Starting with the given partial order $\leq_S$ on $S$, consider partial orders $\leq$ on $S$ satisfying
(1) $s\leq_S t$ implies $s\leq t$; and
(2) $s\leq t$ implies $gs\leq gt$.
The set of such partial orders is itself partially ordered by inclusion and contains $\leq_S$. As usual, every chain in this poset is bounded above by its union, and therefore by Zorn's Lemma there is a maximal partial order $\leq'$. I claim that this is a total order: for otherwise if $s$ and $t$ are incomparable with respect to $\leq'$, we can extend it to an order in which, say, $s\leq'' t$ and more generally $\gamma s\leq'' \gamma t$ for all $\gamma\in G$. (Note that since $s$ and $t$ are $\leq'$ incomparable, so are $\gamma s$ and $\gamma t$, thanks to condition (2); therefore decreeing $\gamma s\leq'' \gamma t$ does not conflict with $\leq'$.) This contradicts the maximality of $\leq'$; therefore $\leq'$ is a total order on $S$. Moreover it satisfies the conditions in the OP.
To answer your other questions: yes the action of $G$ on $S$ is necessarily free, since if $g\neq 1$ then we have $gs>1s$ or $gs<1s$ for all $s$. Any non-trivial (left-)ordered group is torsion-free and therefore infinite. And finally each orbit is in 1-1 correspondence with $G$ since the action is free, so $G$ and $S$ are infinite assuming $G$ is non-trivial.
