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).

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

$\mathcal{NC}:=\{ A \in M_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$.

By a non-conformal matrix, I mean a matrix whose singular values are distinct. (i.e. I allow non-zero singular matrices in $\mathcal{NC}$).

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

$\mathcal{F}$ has two connected components, both homeomorphic to an open half-plane with one point removed.

Indeed, $Ae_1 \in \operatorname{span}(e_1)$ and $A \in \text{SL}_2(\mathbb{R})$ imply that

$$ A=\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix} \, \, \, \text{for some }\, \lambda \neq 0.$$ $A$ is conformal 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

$$\{ 0<\lambda \neq 1\} \times \mathbb{R} \cup \{1\} \times \mathbb{R}\setminus\{0\}.$$

(The second component corresponds to $\lambda <0$.)

Here is what I know about the topology of $\mathcal{NC}$:

Let $\mathcal D=\{ (\sigma_1,\sigma_2) \, | 0 \le \sigma_1 < \sigma_2\}$. Then the map

\begin{align*} \mu: SO_2\times \mathcal D\times SO_2\to \mathcal{NC}\\ (U,\Sigma,V)\mapsto U\Sigma V^T \end{align*}

is a $2$-fold smooth covering map*. (i.e. $\mu(U,\Sigma,V)=\mu(-U,\Sigma,-V)$, and this is the only ambiguity in $U,V$ for a pre-image of a given point in $\mathcal{NC}$.

Since $SO_2 \cong \mathbb{S}^1$, and since after identifying antipodal points in $\mathbb{S}^1 \times \mathbb{S}^1$, we get the $2$-torus $\mathbb{T}^2$ again, it follows that $\mathcal{NC} \cong \mathbb{T}^2 \times D$.

*I am not entirely sure regarding the behaviour at the boundary points where $\sigma_1=0$, but I don't think this creates a serious problem.