I need to describe all moebius transforms $$\phi_A $$ from the extended complex plane in itself, with $$A \in SL(2, \mathbb{R})$$ that map points from the imaginary axis to the imaginary axis.
so basically
$$\phi_A(xi) = \frac{axi + b}{cxi + d} = yi $$
with $$ad-bc = 1$$
How to go on from here? I'm also not sure whether it makes sense to solve the problem algebraically or to think in terms of a composition possible basic transformations (translation, dilation, inversion).
Algebra helps. But rather than the geometric ideas of translations, dilations, inversions, one should simply look at where certain carefully chosen points go.
I'll use the notation $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $a,b,c,d \in \mathbb R$, $ad-bc=1$ as suggested in your question.
The idea is to assume that $\phi_A$ takes the imaginary axis to the imaginary axis and to see what additional algebraic constraints this places on the numbers $a,b,c,d$.
First, I'm sure you know that $\phi_A$ takes the real axis to the real axis. Since the real and imaginary axes intersect in a single point, namely the origin $0$, it follows that $\phi(0)=0$. From this it follows that $\frac{a \cdot 0 + b}{c \cdot 0 + d} = 0$ and so $b=0$. So now we have $A = \begin{pmatrix} a & 0 \\ c & d \end{pmatrix}$ with $ad=1$.
Next, I'll follow your suggestion with $\phi_A(xi)=yi$, but more specifically I'll consider what happens when $x=1$, so $\phi_A(i) = yi$: $$\frac{ai}{ci+d} = yi, \quad y>0 $$ $$ai=dyi-cy $$ so $cy=0$ which implies $c=0$.
So now we have $A = \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}$, $ad=1$. We can therefore let $a$ be a parameter and solve for $d$ to get $$A = \begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix} $$ And that's all there is to be said: that's the general form of an element of $SL(2,\mathbb R)$ that takes the imaginary axis to itself.
