Given,

$A$ be a $2\times 2$ matrix with real entries such that $A^{\text{'}}=\alpha A+1,$

Now let $A=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]$

Now putting the value in $A^{\text{'}}=\alpha A+I$ we get,

$\Rightarrow \left[\begin{array}{ll}a& c\\ b& d\end{array}\right]=\left[\begin{array}{cc}\alpha a+1& \alpha b\\ \alpha c& \alpha d+1\end{array}\right]$

Now equating both side we get,

$a=\alpha a+1\Rightarrow a=\frac{1}{1-\alpha } .........\left(i\right)$

$b=\alpha c .... \left(ii\right)$ and $c=\alpha b ........\left(iii\right)$

Now from equation $\left(ii\right) \& \left(iii\right)$ we get,

$c=0$ or $\alpha =\pm 1$ (not possible as mentioned in question)

$\therefore c=0$

$\Rightarrow c=0, b=0$
Also $d=\alpha d+1\Rightarrow d=\frac{1}{1-\alpha }$

Also given, $\left|A^2-A\right|=4$

$\Rightarrow \left|A\right||A-I|=4$

$\Rightarrow {\left(\frac{1}{1-\alpha }\right)}^2{\left(\frac{1}{1-\alpha }-1\right)}^2=4$

Now let $\frac{1}{1-\alpha }=t$

So, $t^2{\left(t-1\right)}^2=4$

$\Rightarrow t\left(t-1\right)=\pm 2$

$\Rightarrow t=2 \text{or} t=-1$

$\Rightarrow \alpha =\frac{1}{2},2$

So, sum of all possible value will be $\frac{1}{2}+2=\frac{5}{2}$