$$\Sigma(2,F)=\left\{\begin{bmatrix}a&b\\1-a&1-b\end{bmatrix},a,b\in F\right\}$$
$${\rm Aff}(1,F)=\left\{\begin{bmatrix}a&b\\0&1\end{bmatrix},a\neq 0,a,b\in F\right\}$$
First I tried linear transformation:
$\begin{bmatrix}c&d\\e&f\end{bmatrix}\begin{bmatrix}a&b\\1-a&1-b\end{bmatrix}=\begin{bmatrix}ca+d(1-a)&cb+d(1-b)\\ea+f(1-a)&eb+f(1-b)\end{bmatrix}$
I want $eb+f(1-b)=1$ for any $b\in F$. Then it must be $e=f=1$. But I also want $ea+f(1-a)=0$. This is impossible.
Your parametrization of $\Sigma(2,F)$ isn't entirely correct; if $a=b$ then the matrix is singular. And more critically, linear transformation as you've written it (multiplying by some fixed matrix $M$) doesn't preserve the group product; $(MA)(MB)\neq M(AB)$. You probably want to conjugate your matrix $A\in\Sigma(2,F)$ by some matrix M; that is, to look at $MAM^{-1}$. The most straightforward way of doing this is to try and solve $MA=BM$ for $A=\left(\begin{smallmatrix}\alpha&\beta\\ 1-\alpha&1-\beta\end{smallmatrix}\right)\in\Sigma(2,F)$ and $B=\left(\begin{smallmatrix}a&b\\ 0&1\end{smallmatrix}\right)\in\mathrm{Aff}(1,F)$.