The question is, for which $\alpha$ and $\beta$ are the above maps conjugate?
My first thought was to look at the fixed points. So I looked at $z=\alpha z -\frac{1}{z}$. This map will have two fixed points of the form $\pm\frac{1}{\sqrt{\alpha -1}}$ where $\alpha \neq 1$.
The above led me to a knee jerk conclusion that $\beta$ must be a combination of translations and scalars from $\alpha$.
Am I on the right track?
Edit: What I mean conjugate is given some mobius transofmration $g$, we have $g^{-1}fg=h$ where $g(z)=\alpha z-\frac{1}{z}$ and $h(z)=\beta z-\frac{1}{z}.$
Let $f(z)=\alpha z-\frac{1}{z}$ and $g(z)=\beta z-\frac{1}{z}$. Then, assuming my understanding of your question is correct, you want to find $\alpha,\beta\in\mathbb C$ such that $\bar f(z)=g(z)$ for all $z\in\mathbb C$.
I, personally, can't see where your method is going, but that doesn't mean it's wrong. What I would do, though, is, for $z=x+iy$, express $f$ as $u_1(x,y)+iv_1(x,y)$ and $g$ as $u_2(x,y)+v_2(x,y)$ and solve the system $$ \left\{ \begin{align} u_1&=u_2 \\ v_1&=-v_2 \end{align} \right. ,$$ keeping in mind that we want the solution to work for all $x,y\in\mathbb R$.