If I'm correct, to define a hyperbolic translation you need an ideal point $p$ as a source, a different ideal point $q$ as a sink and a length $d \ne 0$ along the geodesic that joins $p$ and $q$. Let's call such a translation $\tau_{pq}^d$. Do $f = \tau_{pq}^d$ and $g = \tau_{rs}^e$ commute?
I think that $f$ and $g$ do not commute unless $\{p, q\} = \{r, s\}$. Here's my reasoning. I've represented the translation in the Poincaré disk model, where it has equation $f(z) = \frac{az+b}{cz+d}$ for $$A_f = \begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix} p-tq & (t-1)pq \\ 1-t & tp-q \end{pmatrix},$$ where $t>0, t\neq1.$ Building the analogue matrix $A_g$ for $g$, with parameter $u >0, u \ne 1$, and assuming $fg=gf$, it must be the case that $A_fA_g = A_gA_f$ (because composition of Möbius transformations is given by the product of their matrices), yielding the equation $$(1-t)(1-u)(rs-pq)=0,$$ which implies $rs=pq$, and also the equation $$(1-t)(1-u)pq[(p+q)-(r+s)] = 0,$$ which implies $r+s=p+q.$ As a result $\{p, q\} = \{r, s\}.$
But there is one subtle problem that could arise: we only know that $A_fA_g = \lambda A_gA_f$ for some $\lambda \in \ \mathbb C^*$ (because multiplying the matrix by a nonzero scalar does not change the transformation). In that case, taking determinants, $\lambda^2=1$ and so $\lambda = \pm 1$. The case $\lambda = 1$ is the previous one. If $\lambda = -1$, then $\operatorname{tr}(A_fA_g) = 0$. The zero trace gives the equation for the cross-ratio $$(p, q, r, s) = \frac{tu+1}{t+u},$$ but the other equations that follow are harder, so I'm not sure if this is a possibility.
Assuming both translations are nontrivial and the source/sink are not both the same, they cannot commute.
Suppose $\tau$ translates from $p$ to $q$ and $\tau'$ translates from $p'$ to $q'$. Then $\tau\tau'=\tau'\tau$ is equivalent to $\tau\tau'\tau^{-1}=\tau'$. But the conjugate $\tau\tau'\tau^{-1}$ translates from $\tau p'$ to $\tau q'$, and $\tau$ translates nontrivially on the ideal boundary - the only ideal points fixed by $\tau$ are $p$ and $q$, so whichever of $p'$ or $q'$ doesn't match $p$ or $q$ respectively is altered by $\tau$, so $\tau\tau'\tau^{-1}$ cannot have the same source/sink as $\tau'$, so cannot be $\tau'$.