Your map sending $(-1,0,1)$ to $(0,i,\infty)$ is given by $i = e^{i\pi/2}$ times the cross ratio $[z,0,-1,1] = \frac{z+1}{z - 1} : \frac{1}{-1}$, so it is $$\tilde{\phi}(z) = e^{i\pi/2}\cdot \frac{1+z}{1-z}.$$
It is easy to see geometrically (or by a direct calculation) that $\tilde\phi(L)$ is the quadrant $\{z:\,\operatorname{Im}{z} \gt |\operatorname{Re}{z}|\}$ since the circles $|z-i|=\sqrt{2}$ and $|z+i|=\sqrt{2}$ are sent to the lines $\{\operatorname{Im}{z} = \operatorname{Re}{z}\}$ and $\{\operatorname{Im}{z} = -\operatorname{Re}{z}\}$. Note that the angles at the vertices of the lune are equal to $\pi/2$.
It is more convenient to work with $\displaystyle\phi(z) = e^{i\pi/4}\cdot \frac{1+z}{1-z}$ which sends $L$ to the quadrant $\{z:\operatorname{Re}{z}, \, \operatorname{Im}{z}\gt 0\}$, then square to get to the upper half plane and apply the Cayley-transfom
$\displaystyle\kappa(z) = \frac{z-i}{z+i}$ to get to the unit disk.
The solution to your problem then is $\kappa((\phi(z))^2)$, which you can compute yourself if needed.
The general procedure is succinctly explained in GEdgar's answer.
We have a sequence of maps
$$f_i:\ D_{i-1}\to D_i,\quad z_{i-1}\to z_i\qquad (1\leq i\leq 5)\ .$$
Here $z_i$ does not denote a certain point in the $z$-plane, but the coordinate variable in the $i$th auxiliary complex plane. $Z_0:=\sqrt{2}-1$ is the $z_0$-coordinate of a certain point $Z$ we are interested in.
$D_0$ is the unit disk in the $z_0$-plane minus the points $z_0\leq0$. The map
$$f_1:\ z_0\mapsto z_1:={\rm pv}\sqrt{z_0}$$
maps $D_0$ onto the right half $D_1$ of the unit disk in the $z_1$-plane. Thereby the point $Z_0$ is mapped onto a point $Z_1\in\ ]0,1[\ $.
The Moebius map
$$f_2:\ z_1\mapsto z_2:=-{z_1-i\over z_1+i}$$
maps $i$ to $0$ and $-i$ to $\infty$. Furthermore $f_2(0)=1$, $\ f_2(1)=i$. From general properties of Moebius maps it then follows that $D_2:=f_2(D_1)$ is the first quadrant, and that $f_2$ maps the real axis onto the unit circle. Therefore $Z_2=f_2(Z_1)$ is a point between $1$ and $i$ on the unit circle.
The map
$$f_3:\ z_2\mapsto z_3:=z_2^2$$
maps the first quadrant $D_2$ onto the upper half-plane $D_3$, whereby $f_3(1)=1$, $\ f_3(i)=-1$, and the quarter unit circle in $D_2$ is mapped onto the upper half of the unit circle in $D_3$. Therefore the point $Z_3:=f_3(Z_2)$ is lying on this upper half of the unit circle, too.
The Moebius map
$$f_4:\ z_3\mapsto z_4:=i{z_3-i\over z_3+i}$$
maps the upper half plane $D_3$ onto the unit circle $D_4$. Thereby $f_4(-1)=-1$, $\ f_4(1)=1$, and the unit circle of the $z_3$-plane is mapped onto the real axis of the $z_4$-plane. It follows that $Z_4=f_4(Z_3)$ is a real number between $-1$ and $1$.
Doing the calculations $Z_4$ should simplify to an expression defining a real number $\alpha\in\ ]{-1},1[\ $. Letting
$$f_5:\ z_4\mapsto{z_4-\alpha\over 1-\alpha z_4}$$
you finally arrive at the required map
$$f:=f_5\circ f_4\circ f_3\circ f_2\circ f_1\ .$$
Best Answer
Hint
Consider this three transformations
$T_1(z)=iz,$ which maps from $\{z\in \mathbb{C} :|z|>1, Re(z)>0\} $ to $\{z\in \mathbb{C} :|z|>1, Im(z)>0\}.$
$T_2(z)=\frac 12(z+\frac 1z),$ which maps from $\{z\in \mathbb{C} :|z|>1, Im(z)>0\} $ to the upper half plane.
and
$T_3(z)=\frac{z-i}{z+i},$ which maps from the upper half plane to the unit circle.
and take $T=T_3\circ T_2\circ T_1$