Consider the hyperbolic matrices
$$
A = 
\begin{pmatrix}
2 & 1 \\
1 & 1
\end{pmatrix}
\quad
\mbox{and}
\quad
b = 
\begin{pmatrix}
2 & -1 \\
-1 & 1
\end{pmatrix}
$$
Working in the upper half plane model of $\mathbb{H}^2$, we take $a$ to be the (oriented!) geodesic from $0$ to $-1$ and $a'$ to be the geodesic from $1$ to $\infty$.  Similarly, take $b$ to be the geodesic from $0$ to $1$ and $b'$ to be the geodesic from $-1$ to $\infty$.  (It helps to draw a figure at this point.) Then $A$ takes $a$ to $a'$ and $B$ takes $b$ to $b'$, all preserving orientations.  Also, the axis of $A$ is transverse to (but not perpendicular to) $a$ and $a'$; similarly the axis of $B$ is transverse to $b$ and $b'$.  

We deduce that $A$ and $B$ generate a free rank two subgroup of $\mathrm{SL}(2, \mathbb{Z})$.  However, their commutator $ABA^{-1}B^{-1}$ is parabolic.  So, to answer the original question, we instead consider the subgroup generated by $A^2$ and $B^2$.  A standard "ping-pong" argument shows that these generate a free group of rank two where all non-identity elements are hyperbolic.