If a Möbius transformation
$$\phi: z \mapsto \frac{a z + b}{c z + d}$$ preserves the upper half-plane, then by continuity (as a self-map of the Riemann sphere) it must map $\mathbb{R} \cup \{\infty\}$ to itself, and some algebra forces that (possibly after canceling a common nonreal factor) $a, b, c, d$ are all real. Any such map maps the u.h.p. either to itself or to the l.h.p., and it does the former iff $a d - b c > 0$. Like you say, direct checking shows that any such transformation preserves the Poincare metric, and also the orientation of the upper half-plane.
On the other hand, a typical Möbius transformation that maps the u.h.p. to the unit disk is $$z \mapsto \frac{z - i}{z + i},$$
and no such transformation has all real coefficients.
Remark For any invertible complex matrix $A := \begin{pmatrix}a & b\\c & d\end{pmatrix}$ and $\lambda \in \mathbb{C}^*$, the matrix $\lambda A = \begin{pmatrix}\lambda a & \lambda b\\ \lambda c & \lambda d\end{pmatrix}$ determines the Möbius transformation
$$
\phi: z \mapsto \frac{\lambda a z + \lambda b}{\lambda c z + \lambda d} = \frac{a z + b}{c z + d} ,
$$
namely the same transformation determined by $A$, but some algebra shows that this is the only redundancy, so we can think of the group of Möbius transformations as $GL(2, \mathbb{C}) / \mathbb{C}^* \cong PSL(2, \mathbb{C})$. The above shows that the group of (oriented) isometries of the Poincare half-plane is the proper subgroup $PSL(2, \mathbb{R})$.
$$ {\cosh ^{-1}} \frac{(R+ 1/R)}{2} $$
$$= {\cosh ^{-1}} \frac{ e^ {\log (R)}+ e^{-\log(R)}}{2} $$
$$ = {\cosh ^{-1}} [ \cosh (\log R)] = \log R . $$
EDIT1:
BTW, why do we assume unit (abs value) Gauss curvature? Should it not appear symbolically at least in a formula ?
$$ \operatorname{dist} (\langle a, r \rangle, \langle a, R \rangle) = \operatorname{ a \cdot \,arcosh} \left( 1 + \frac{ {(R - r)}^2 }{ 2 r R } \right) $$
Best Answer
The Poincare disk and upper half-plane models are related by a Möbius transformation that maps the disk to the plane. Such a transformation has the form $$ f(z) \;=\; \frac{az+b}{cz+d} $$ for some constants $a,b,c,d\in\mathbb{C}$ such that $ad-bc\ne 0$.
In my mind, the "usual" Möbius transformation $f$ satisfies $f(-i) = 0$, $f(1) = 1$, $f(-1)=-1$, and $f(i)=\infty$, though there may be other conventions. Solving for the coefficients gives the formula for $f$: $$ f(z) \;=\; \frac{z + i}{iz+1}. $$ The inverse of this maps the upper half-plane to the disk, and is given by $$ f^{-1}(z) \;=\; \frac{z-i}{-iz+1} $$
Note For the specific requirements you gave for $f$ where $f(0) = i$, $f(-i)=\infty$, and $f(i) = 0$, you would want the $180^\circ$ rotation of the function $f$ above, i.e. $f(z) = \dfrac{z-i}{iz-1}$.