My try : I selected two arbitrary points $(x_e,y_e)$ and $(x_h,y_h)$ and I computed the distance which I say to be minimum.
$y_e$ can be eliminated (expressed as a function of $x_e$ since the point is along the ellipse).
$y_h$ can be eliminated (expressed as a function of $x_h$ since the point is along the hyperbole).
Now, I want that the derivatives of the distance with respect to both $x_e$ and $x_h$ be equal to zero. This leads to a terrible system of two equations for the two unknowns $x_e$ and $x_h$ but, fortunately, we only need to consider the first quadrant because of the symmetry. Moreover, complex solutions can be discarded.
I cannot get an analytical solution but using numerical methods, the only solution corresponds to $(1.62723,0.581406)$ along the ellipse and $(2.39098,1.67296)$ along the hyperbole.
Using Lagrange multipliers (then $6$ variables) took me to the same result.
The results were later confirmed building a contour plot of the distance as a function of parameters $x_e$ and $x_h$.
Added later to this answer
The problem can reduce to a single variable (say $x_e$) problem, eliminating $x_h$ as a function of $x_e$ from one of the derivative of the distance with respect to the variables. What is then left is to solve the remaining derivative for $x_e$ (the advantage of keeping $x_e$ as the single variable of the problem is that it is bounded between $0$ and $2$).
So, the problem is now reduced to the solution of a single equation for a single unknown. This works and leads to identical results. Newton methods works quite well except if the iterations start at $x_e=0$ or $x_e=2$ which correspond to infinite branches. But, for example, starting iterations at $x_e=1$, the successive iterates are $1.32085$, $1.57018$, $1.62728$, $1.62723$.
This approach could be used for finding the closest distance between two conics.