Here's a proof that avoids covering space arguments, using a bit of differential topology, and using some surface topology arguments including the (smooth) Schönflies Theorem.
One can cut $S$ along a simple closed curve $\gamma$ (representing $[a,b]=[c,d]^{-1}$) into two compact subsurfaces $S_0,S_1$ with boundary, each a one-holed torus, each with fundamental group free of rank 2, so that the inclusion map $S_0 \hookrightarrow S$ takes a free basis of $\pi_1(S_0)$ to the two group elements $a,b$, so that the other inclusion map $S_1 \hookrightarrow S$ takes a free basis of $\pi_1(S_1)$ to $c,d$, and so that $\gamma = \partial S_0 = \partial S_1$ represents infinite order elements of $\pi_1(S_0)$ and of $\pi_1(S_1)$, namely $[a,b]$ and $[c,d]^{-1}$ respectively.
It suffices to prove that the inclusion map $S_0 \hookrightarrow S$ induces an injection on fundamental groups.
Since the inclusion of $\text{interior}(S_0)$ into $S_0$ is a homotopy equivalence, it suffices to prove that $\text{interior}(S_0) \hookrightarrow S$ induces an injection on fundamental groups. And for this, it suffices to prove that for every continuous closed curve $f : S^1 \to \text{interior}(S_0)$, if $f$ extends to a continuous function $F : D^2 \to S$ then $f$ also extends to a continuous function $F_0 : D^2 \to \text{interior}(S)$. By approximation methods in differential topology, we may assume that $f$ and $F$ are both smooth.
Now we apply transversality methods of differential topology. We know that $F^{-1}(\gamma) \subset \text{interior}(D^2)$, and so (applying perturbation methods in differential topology) we may assume that the map $F$ is tranverse to the curve $\gamma$. The subset $F^{-1}(\gamma) \subset \text{interior}(D^2)$ is therefore a pairwise disjoint collection of smooth, simple closed curves in the interior of $D^2$. Let $n$ be the number of those curves, denoted $C_1,...,C_n$.
By applying the Schönflies Theorem, each $C_i$ subdivides $D^2$ into two pieces: $\text{inside}(C_i)$ is diffeomorphic to $D^2$ with boundary $C_i$; and $\text{outside}(C_i)$ is diffeomorphic to an annulus with boundary $C_i \cup S^1$. Pick a collar neighborhood of $C_i$, parameterized as $C_i \times [-1,+1] \to D^2$ so that $C_n \times \{0\} \to C_i$ and $C_i \times \{+1\}$ is a simple closed curve $C'_i \subset \text{outside}(C_i)$.
The proof now proceeds by induction on $n$. The relation $C_i \subset \text{inside}(C_j)$ is a partial ordering on the finite set $\{C_1,\ldots,C_n\}$, and so there is a minimal element. Let's permute the notation so that $C_n$ is a minimal element, hence $\text{inside}(C_n)$ contains no other $C_i$. It follows that $F(\text{inside}(C_n)) \subset S_j$ for $j=0$ or $1$, and therefore $F | C_n $ is null homotopic in $S_j$. But $F | C_n \subset \gamma$ and $\gamma$ represents an infinite order element of $\pi_1(S_j)$, and so it follows that $F | C_n$ is null homotopic in the circle $\gamma$ itself. From this it follows that $F \mid C'_n$ is null homotopic in $\text{interior}(S_k)$ (namely in the complement of $S_j$, i.e. $k=1-j$). We can therefore redefine $F \mid \text{inside}(C'_n)$ to be a null homotopy of $F \mid C'_n$ in $\text{interior}(S_k)$. This redefinition operation removes $C_n$ from $F^{-1}(\gamma)$ with affecting the remainder of $F^{-1}(\gamma)$, thus reducing the number of components of $F^{-1}(\gamma)$, and completing the induction step.
