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.
I poked around in a few standard references, but did not find this exact statement. However it is "easily" deduced from material in various places. For example, you may enjoy reading Chapter 3 of Noneuclidean tesselations and their groups by Wilhelm Magnus. (Note the amazing collection of illustrations, mostly taken from the works of Fricke and Klein, starting on page 159.) In a somewhat different direction, you could use the "combination theorem" of Klein-Maskit. See Section VIII.A.1 of Kleinian groups by Bernard Maskit.