Can two elliptic isometries generate a free group?

Here’s my fairly extended example:

If you’re dealing with two centers of rotation, you might as well put them both on the imaginary axis (in Upper Half Plane model), and one of them at $i$, the other at $Ki$ with $K>1$. In case you needed to know, the distance then would be $\log K$. I made another decision, to do the contrary of what you suggested, and take both rotations to be of $90^\circ$ CCW. To make things easier, I’ll prefer the matrix representation: $$ \text{about $i$:}f(z)=\frac{z+1}{-z+1}\>, \text{i.e.} \begin{pmatrix}1&1\\-1&1\end{pmatrix}\>;\qquad\text{about $Ki$:}g=\begin{pmatrix}K&K^2\\-1&K\end{pmatrix}\,. $$ Then we get $$ f\circ g=\begin{pmatrix}K-1&K^2+K\\ -K-1&K-K^2\end{pmatrix}\,,\quad g\circ f=\begin{pmatrix}K-K^2&K^2+K\\-K-1&K-1\end{pmatrix}\,. $$ In both cases, the condition to be hyperbolic is $K^2-6K+1>0$; the roots of the quadratic are $K=3\pm\sqrt8$, so for an explicit example, I chose $K=6$. Here, the condition for $(f\circ g)(z)=z$ is $0=7(z^2+5z+6)=7(z+2)(z+3)$ so that the fixed points of the hyperbolic $f\circ g$ are $-2,-3$, while the fixed points of $g\circ f$ are $2,3$. Thus if what you say about hyperbolic with different fixed-point sets is correct, there’s a rank-two free group sitting in $\langle f,g\rangle$