Yes, it's easy to interpolate. Let $r(z) = f^{-1}g(z) = \frac{r\_1z + r\_2}{r\_3z + r\_4}$, where $r\_1r\_4 - r\_2r\_3 = 1$. Diagonalize the matrix (if possible): $$R = \begin{bmatrix}r\_1 & r\_2 \\\\ r\_3 & r\_4\end{bmatrix} = V\begin{bmatrix}\lambda & 0 \\\\ 0 & \lambda^{-1}\end{bmatrix}V^{-1} = V\begin{bmatrix}e^{c} & 0 \\\\ 0 & e^{-c}\end{bmatrix}V^{-1}.$$ Let $$S(t) = V\begin{bmatrix}e^{ct} & 0 \\\\ 0 & e^{-ct}\end{bmatrix}V^{-1} = \begin{bmatrix}s\_1(t) & s\_2(t) \\\\ s\_3(t) & s\_4(t)\end{bmatrix}$$ for $0 \leq t \leq 1$ (so $S(0) = I$ and $S(1) = R$), and let $$s(z,t) = \frac{s\_1(t)z + s\_2(t)}{s\_3(t)z + s\_4(t)}$$ (so $s(z,0) = z$ and $s(z,1) = r(z) = f^{-1}g(z)$) and finally define $h(z,t) = f(s(z,t))$, so $h(z,0) = f(z)$ and $h(z,1) = f(f^{-1}g(z)) = g(z)$. It seems like a lot of steps, but conceptually, it's very simple. By postcomposing by the inverse of one of your transformations, you effectively reduce the problem to the case where one transformation is the identity. Then it's clear how to interpolate: embed the other transformation in a one-parameter subgroup of the isometry group of the hyperbolic plane, which diagonalizing makes easy. (Exercise: show that $s(s(z,t\_2),t\_1) = s(z,t\_1+t\_2)$, so $t\mapsto (z\mapsto s(z,t))$ is indeed an embedding of $\mathbb{R}$ into the group of hyperbolic isometries (i. e., a one-parameter subgroup).)
If the matrix $R$ is not diagonalizable, then you can repeat the above construction using the Jordan canonical form instead:
$$R = V\begin{bmatrix}1 & 1 \\\\ 0 & 1\end{bmatrix}V^{-1},$$ and use $$S(t) = V\begin{bmatrix}1 & t \\\\ 0 & 1\end{bmatrix}V^{-1}.$$