A Möbius transformation that maps the upper half plane to itself must map the boundary, i.e. the real line, to itself as well. With the standard definition for the transformation,
$$f(z) = \frac{az+b}{cz+d},$$
it is fairly straightforward to see that the parameters $a, b, c, d$ must be real to map the real line to itself (try plugging in $z=0$, $z=\infty$, and then one other real number).
Then since $f$ is supposed to interchange $z_1$ and $z_2$, that means applying $f$ twice should send $z_1$ back to itself, and the same for $z_2$. Note that $f(f(z))$ is again another Möbius transformation with real coefficients. You can now solve the equation for the fixed points of the transformation (or look up the formula). This should answer the question about whether the transformation is possible.
For your first example, because the boundary of the upper half-plane is a "circle" (in the Riemann sphere sense (sorry, Riemann sphere, not Bloch sphere)), and the boundary of the unit disk is a circle (plainly, but also in the Riemann sphere sense), we try to map the boundary of the one to the boundary of the other. We start with a generic Möbius transformation:
$$w=\frac{az+b}{cz+d}.$$
Here $a,b,c,d\in\mathbb{C}$ and $ad-bc\not=0.$
Let's remember that $z$ is a point in the upper half-plane, and $w$ is in the unit circle. So if we consider the following table:
$$\begin{array}{c|c}z &w \\ \hline 0 &b/d \\ \hline 1 &(a+b)/(c+d) \\ \hline \infty &a/c \end{array},$$
we need all three points in the second column to be unique points on the unit circle. How did I get the points for $z?$ The answer is this: they're the most straight-forward points on the real axis (the boundary of the upper half plane) to plug into the transformation. We can see right away that we need $|b|=|d|,\; |a|=|c|,$ and $|a+b|=|c+d|,$ in order for the $w$ points to be on the unit circle. There are a number of ways to do this, so let's try $a=b=c=d=1:$
$$\begin{array}{c|c|c}z &w &\text{actual} \; w \\ \hline 0 &b/d &1 \\ \hline 1 &(a+b)/(c+d) &1 \\ \hline \infty &a/c &1 \end{array}.$$
Clearly, this won't work: the $w$ have to be unique! Could we work backwards? How if we could map $0\to 1, \; 1\to i, \; \infty\to-1?$ Are there $a,b,c,d$ to make that work? Let's see:
$$\begin{array}{c|c|c}z &w &\text{actual} \; w \\ \hline 0 &b/d &1 \\ \hline 1 &(a+b)/(c+d) &i \\ \hline \infty &a/c &-1 \end{array}.$$
So here $b=d$ and $a=-c$. To get $i$ out of $\dfrac{a+b}{c+d}=\dfrac{a+b}{-a+b}$ would require $a+b=i(-a+b),$ so (remembering that $a,b\in\mathbb{C}$) we would have
\begin{align*}
a(1+i)&=b(i-1) \\
a&=\frac{b(i-1)}{i+1}.
\end{align*}
We have one degree of freedom here, now. Let's just let $b=1,$ and see what falls out. We get
$$a=\frac{i-1}{i+1}\cdot\frac{i-1}{i-1}=\frac{(i-1)(i-1)}{2}=\frac{-1-2i+1}{2}=-i.$$
Then, from before, we have $d=1$ and $c=i,$ so the mapping is
$$w=\frac{-iz+1}{iz+1}.$$
We can see that $ad-bc\not=0.$ What about the interior? Let's see if $z=i$ maps to the interior of the unit circle. $z=i$ goes to $(1+1)/(0)$. Oops! Can't do that. We can see that $z=-i$ maps to $w=0$. So what's going on? We actually mapped the lower half-plane to the unit circle! The upper half-plane got mapped to everything outside the unit circle. What went wrong? Well, we had some freedom in choosing what to map $0,1,\infty$ to, so we must have done that wrong. Let's try going around the boundary of the unit circle the other way ($0,1,\infty$ got mapped to a counter-clockwise orientation on the unit circle before), and see if we can fix it:
$$\begin{array}{c|c|c}z &w &\text{actual} \; w \\ \hline 0 &b/d &-1 \\ \hline 1 &(a+b)/(c+d) &? \\ \hline \infty &a/c &1 \end{array}.$$
We'll leave the middle one blank for now, but to get the clockwise orientation, it'll need to have a positive imaginary part. Let's try $a=c=1$. Now $b=-d,$ and we're going to need nonzero imaginary parts to get that third point right. The easiest thing to try is $b=i, \; d=-i$. Let's check the middle row:
$$\frac{a+b}{c+d}=\frac{1+i}{1-i}\cdot\frac{1+i}{1+i}=\frac{1+2i-1}{2}=i, $$
which makes our transformation
$$w=\frac{z+i}{z-i}.$$
Does this work? We check the interior again: $z=i$ should map to the interior of the unit circle. We get $w=(2i)/(0),$ which doesn't work again! So we try it the other way, with $b=-i$ and $d=i$, and you'll find that it works.
I wanted to show you how I think about Möbius transformations, and how I get one to work. It basically all comes down to this: if you generalize the concept of circle to circles or lines (lines still being a circle on the Riemann sphere), then Möbius transformations map circles to circles. Use that to map one boundary to the other, and then get the interiors to match up. That's the basic plan of attack.
Best Answer
A common approach it to find a “conjugate” map which is simpler to investigate. Here we can conjugate $L$ with $$ T(z) = \frac{z-\alpha}{1-\overline{\alpha}z} $$ so that the fixed point is transformed to the origin.
If $L$ is a Möbius transformation mapping the unit disk onto itself with $L(\alpha) = \alpha$ then $$ S = T \circ L \circ T^{-1} $$ is a Möbius transformation mapping the unit disk onto itself with $S(0) = 0$ and $S'(0) = L'(\alpha)$.
It follows that $S(z) = e^{i\theta} z$, and therefore $$ L(z) = T^{-1}(e^{i\theta} T(z)) \, . $$