Is this subset of matrices contractible inside the space of non-conformal matrices? 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.
 A: Edited. In the first version of the answer I was assuming that the space in which the contraction was taking place was not $\cal NC$ but the complement to non-conformal matrices in $SL(2,\mathbb R)$. I'll suggest a fix for this now.
Note, that we have a natural continuous map $u: {\cal NC}\to S^1=\mathbb RP^1$. Namely, to each matrix $A$ from  ${\cal NC}$ we can associate the following one-dimensional subspace $u(A)\in \mathbb R^2$. Take the matrix $AA^{*}$ and take the eigenspace corresponding to the maximal eigenvalue of $AA^*$ (there will be two distinct eigenvalues since $A$ is not conformal). 
So, if we find a closed path $\gamma$ in $\cal F$, such that its image $u(\gamma)\subset S^1$ is not contractible, we are done. How to find such a path is explained in the previous answer to this question, which the path $\gamma(t)$ constructed in the previous answer below. ( I believe that what I suggest works for several reasons but I don't have time to work out all the details now. By the way, it is also funny that $\pi_1(\cal NC)$ seem to be equal $\mathbb Z^2$, moreover it deformation retracts to $T^2$, I believe.)
Previous answer.
It is not contractible. Let us associate to each matrix $A\in SL_2(\mathbb R)$ the following vector $v(A)$. Take an orthogonal matrix $O\in SO_2(\mathbb R)$ such that $OA(e_1)$ is proportional to $e_1$ with a positive coefficient. Then set $v(A)=OA(e_2)$. We get a map to the upper half plane:
$$V:SL(2,\mathbb R)\to \{y>0\}$$
Note that the image of confromal matrices is the point $(0,1)$, and the image of any component $\cal F$ is the complement to $(0,1)$. And so each component can be identified with this puncutred half-plane.  Hence it is enough to construct a path in $\cal F$ whose image under $V$ is not contractible in $\{y>0\}\setminus \{(0,1)\}$. This is easy, just take a non-contractible path $\gamma(t)\subset \{y>0\}\setminus \{(0,1)\}$ (that has a non-zero winding number around $(0,1)$),  and consider the unique path of matrices $A_t\subset \cal F$ such that $A_t(e_2)=\gamma(t)$.
A: Yes, they are contractible in $\mathcal{NC}$, in fact even with $SL$ rather than $GL$.  First simultaneously rotate $\mathcal{F}$, so the typical element is sent to
$$
\begin{pmatrix} \cos a & -\sin a \\\ \sin a & \cos a \end{pmatrix}\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix}=\begin{pmatrix} \lambda \cos a & y\cos a-\lambda^{-1}\sin a \\\ \lambda \sin a & y\sin a+\lambda^{-1}\cos a \end{pmatrix}
$$
for some small $a$.  These are not conformal, since the $(2,1)$-entry is nonzero. 
 Then simultaneously shrink all these matrices down to the $\lambda=\pm 1$, $y=0$ matrix (depending on the connected component),
$$
\pm\begin{pmatrix} \cos a & -\sin a \\\ \sin a & \cos a \end{pmatrix}.
$$
