If $GL_2(\mathbb{R})\times\mathbb{R}^2\to \mathbb{R}^2$ satisfy $(g,x)\mapsto gx$. $\ $ Suppose $H_x=\{g\in G:gx=x\}$. Can we prove $H_x\cong H_1\rtimes H_2$,$\ $ $H_1, H_2$ are Lie subgroups and $H_1\cong G_{a/\mathbb{R}},\ H_2\cong G_{m/\mathbb{R}}$?
Here $G_{a/\mathbb{R}}=(\mathbb{R},+)$, $G_{m/\mathbb{R}}=(\mathbb{R}\setminus\{0\},\times)$ and $$ H_1\rtimes H_2:=(H_1\times H_2,*):(a,b)*(c,d):=(a\phi_b(c),bd)$$ where $\phi_b \in Aut(H_1)$.
First note that $GL_2(\mathbb{R})$ acts transitively on $\mathbb{R}^2\backslash \{0\}$ then all stabilizers of non zero elements are conjugate to each other. So let us fix $x= \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. It follows by direct computation that $$ H_x = \left\{\begin{pmatrix} 1 & a \\ 0 & b \end{pmatrix} \mid a, b\in \mathbb{R}, b \neq 0 \right\} $$ Then we define $$ H_1 = \left\{\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} \mid a\in \mathbb{R} \right\}, \quad H_2 = \left\{\begin{pmatrix} 1 & 0 \\ 0 & b \end{pmatrix} \mid b\in \mathbb{R}\setminus \{0 \}\right\}, $$
Note that for any element of $H_x$ we can write it as a product $$ \begin{pmatrix} 1 & a \\ 0 & b \end{pmatrix} = \begin{pmatrix} 1 & \frac{a}{b} \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & b \end{pmatrix} $$ Hence we define map of sets $\Psi \colon H_x \rightarrow G_a \times G_m$ by $$ \Psi \left(\begin{pmatrix} 1 & a \\ 0 & b \end{pmatrix}\right) = \left(\frac{a}{b}, b\right) $$ We may define an operation on $G_a \times G_m$ that makes $\Psi$ a homomorphism. It follows from $$ \begin{pmatrix} 1 & a \\ 0 & b \end{pmatrix} \cdot \begin{pmatrix} 1 & c \\ 0 & d\end{pmatrix} = \begin{pmatrix} 1 & c+ad \\ 0 & bd \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{b}\frac{c}{d} + \frac{a}{b} \\ 0 & 1 \end{pmatrix}\cdot \begin{pmatrix} 1 & 0 \\ 0 & bd \end{pmatrix} $$ that the operation is $$ (p,q)\ast (r,s) = (p+q^{-1} r , qs) $$ i.e. the automorphism is $\phi_q(r) = q^{-1}r$.
It only remains to show that $\Psi$ is a group isomorphism.