The answer is yes. It's a little easier if we consider oriented geodesics (which is fine for your argument). Let $\alpha$ and $\beta$ be two oriented simple closed curves on $S_g$ that intersect once. Let $a$ and $b$ be geodesics that are homotopic to $\alpha$ and $\beta$, respectively. Then $a$ and $b$ only intersect once, say at a point $x \in S$. Now let two isometries $f_1,f_2 : S \rightarrow S$ be given. Assume that $f_1(a) = f_2(a)$ and $f_1(b) = f_2(b)$ (as oriented geodesics). I claim that $f_1=f_2$. Indeed, it is clear that $f_1(x)=f_2(x)$. Moreover, if $v$ is the unit tangent vector at $x$ pointing in the direction of $a$ (remember, $a$ has an orientation), then we also know that $f_1(v)=f_2(v)$. Since an element of $SO(2)$ is determined by its value at a single non-zero vector, this implies that both the values and the derivatives of $f_1$ and $f_2$ agree at $x$, and a standard result of Riemannian geometry (using the exponential map) says that then $f_1 = f_2$.
This implies that you can take $S$ to be the set of all geodesics whose lengths are at most the lengths of $a$ or $b$.