Consider the upper-half plane model of the hyperbolic plane $\mathbb {H}^2.$
Now consider two lines in it given as $\ell_1:=\lbrace { (x, y)\in \mathbb {H}^2 \vert x^2 +y^2=r^2\rbrace}, \ell_2:=\lbrace { (x, y)\in \mathbb {H}^2 \vert x^2 +y^2=R^2\rbrace} $, where $0 <r <R.$
I would like to know the (minimal) distance between the lines and those two points $z*, w*$ on $\ell_1, \ell_2 $ which have the least distance, i.e. such that $d ( z*, w* )\leq d ( z, w) , \forall z\in\ell_1, w\in\ell_2. $
My suggestion is that the minimal distance should be between the points $(0, R), (0, r) $, but I could not prove it yet.
By now, I have computed for any fixed point $z=r_1 (\cos\theta, \sin \theta)\in\ell_1$ the unique point $w=w (\theta)\in\ell_2$ which has the minimal distance to z and computed the distance between them. But the formulas are so lengthy and complicated that it is not easy to compare the distances.
I hope there is a more elegant way or argument to investigate the distances.
Best wishes

Two hyperbolic lines that don't intersect and are not parallel (that is, they are ultra parallel) do have a unique common perpendicular. The common perpendicular intersects the two hyperbolic lines at two points that are the closest points between the two lines. The task is to find the common perpendicular.
Since the Poincaré upper half plane model is conformal, angles seen by the Euclidean eye are actually the hyperbolic angles.
It can be seen clearly in the following figure that the green vertical line (hyperbolic straight) is perpendicular to both the red and blue circles (hyperbolic straights). That is, the green line is the common perpendicular.
This common perpendicular intersects the circles (hyperbolic straights) at $(0,r)$ and $(0,R)$. So, we have to calculate the distance between these points as the OP conjectured.
The distance formula in the Poincare half plane model can be applied:
$$d_{\text{min}}=\operatorname{arcosh}\left(1+\frac{(R-r)^2}{2rR}\right).$$