Set $\mathcal{F}:=\{ A \in \text{SL}_2(\mathbb{R}) \, | \, Ae_1 \in \operatorname{span}(e_1) \, \, \text{ and } \, \, A \, \text{ is not conformal} \,\}$, and

$\mathcal{NC}:=\{ A \in \text{GL}_2^{+}(\mathbb{R}) \, | A \, \text{ is not conformal} \,\}$.

Is each connected component of $\mathcal{F}$ contractible in $\mathcal{NC}$?

$\mathcal{F}$ has two homeomorphic connected components, both not contractible on their own; indeed, $Ae_1 \in \operatorname{span}(e_1)$ and $A \in \text{SL}_2(\mathbb{R})$ imply that

$$ A=\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix} \, \, \, \text{for some }\, \lambda \neq 0.$$ Such a matrix is conformal (isometric) if and only if $\lambda=\pm 1$ and $y=0$. So, $A$ is not conformal if f $\lambda \neq 1,-1,0$ or $\lambda=\pm 1$ and $y \neq 0$. Thus, one connected component of $\mathcal{F}$ is homeomorphic to

$$=\{(\lambda,y) \, |\, 0<\lambda \neq 1 \} \cup \{(\lambda,y) \, |\, \lambda = 1, y \neq 0 \} =$$ $$\{ 0<\lambda \neq 1\} \times \mathbb{R} \cup \{1\} \times \mathbb{R}\setminus\{0\}.$$

(The second component corresponds to $\lambda <0$ and is homeomorphic to the first).