The answer to the question *as stated* is "no": take for $S$ the plane $\mathbb{C}$ with the metric induced by the differential $z\mathrm{d}z$; then $S^{\ast }=\mathbb{C}\smallsetminus \{0\}$ and there is simply no geodesic from $1$ and $\mathrm{i}$. Or, if you want closed loops and free homotopy, there is no geodesic in the free homotopy class of the loop around $0$. Intuitively, it is clear what happens: if you draw a curve and try to pull it straight, you are forced to go through the cone point at the origin.

If, however, you consider the surface $S$ itself and extend the definition of a geodesic so as to work on Euclidean metrics with cone points as well, the answer is "yes" *if $S$ is complete as a metric space*. The usual definition of geodesics in this context is one which works for all metric spaces: locally isometric maps from an interval. The statement you asked for is then a consequence of the Arzelà-Ascoli theorem.

These things are treated in detail in the textbook *Quadratic differentials* by Kurt Strebel.