This very special case follows from "general principles" (namely a version of the ping-pong lemma) but it is also possible to give a direct proof, as follows.
Suppose that $a$ and $b$ are the distinct points at infinity fixed by the two parabolic subgroups $A$ and $B$ (note change of notation). Since $\mathrm{SL}(2, \mathbb{R})$ is three-transitive, we can conjugate $\Gamma$ and so assume that $a = \infty$ and $b = 0$. We deduce that elements of $A$ now have the form
$$
\begin{pmatrix}
1 & r\\
0 & 1
\end{pmatrix}
$$
while elements of $B$ have the form
$$
\begin{pmatrix}
1 & 0\\
s & 1
\end{pmatrix}
$$
Taking an inverse if needed, we obtain such elements of $A$ and $B$ where $r$ and $s$ are non-zero and have the same sign. We now multiply to find
$$
\begin{pmatrix}
1 & r\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
s & 1
\end{pmatrix}
=
\begin{pmatrix}
1 + rs & r\\
s & 1
\end{pmatrix}
$$
This has trace $2 + rs > 2$ so is hyperbolic, as desired.
Hmm. Now that I write this, I realise that there is a third proof using the classification of fixed points of isometries, and the intermediate value theorem. It helps to draw a picture and think dynamically.
