I have the following Lie group:
$$G:=\bigg\{\begin{pmatrix} x&0\\y &1/x\end{pmatrix}\bigg\vert x, y\in\mathbb{R} \land y\neq0\bigg\} $$
Now I should calculate the corresponding Lie algebra....
Obviously, $G$ is a subgrouo of the general linear group $\mathrm{GL}_{2}(\mathbb{R})$.....But i have no idea how to construct the Lie algebra, which is the set of all left invariant vector fields.
Two parameters, so two generators, so one commutation relation. Has to be the affine group in one dimension, here transposed. $$ G= \begin{bmatrix} x & 0 \\ y & 1/x \end{bmatrix} $$ goes to the identity for $x=1, y=0$. At the identity, then, you have $$ \partial G/\partial x \to \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \equiv a ~, $$ and $$ \partial G/\partial y \to \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \equiv b ~, $$ so, then, $$ [b,a]=2b . $$
Exponentiating an arbitrary linear combination of these two Lie algebra elements yields the generic form G you started with, since the relevant CBH expansion sums elegantly to a simple closed form: your starting point.
Consider the wisecrack $$ G= \begin{bmatrix} x & 0 \\ y & 1/x \end{bmatrix} = \begin{bmatrix} x & 0 \\ 0 & 1/x \end{bmatrix} \begin{bmatrix} 1 & 0 \\ y x & 1 \end{bmatrix} = e^{(\ln x) ~a} e^{xy~ b} , $$ and its CBH contraction, $$ =\exp \left ( \ln x ~a -2xy\frac {\ln x}{1-x^2} b \right ) = e^{\omega(a + b y/\!\sinh \omega)}, $$ upon the definition $x\equiv e^\omega$.