Continuous at $x=1,3$

$f\left(1^{-}\right)=f\left(1\right) \Rightarrow ae+b{e}^{-1}=c ...\left(1\right)$

$f\left(3\right)=f\left(3^{+}\right) \Rightarrow 9c=9a+6c \Rightarrow c=3a ...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$b=ae(3-e) ...\left(3\right)$

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{cc}a{e}^x-b{e}^{-x},& -1<x<1\\ 2cx,& 1<x<3\\ 2ax+2c,& 3<x<4\end{array}\right.$

$f^{\text{'}}\left(0\right)=a-b, f^{\text{'}}\left(2\right)=4c$

Given $f^{\text{'}}\left(0\right)+f^{\text{'}}\left(2\right)=e$

$a-b+4c=e ...\left(4\right)$

By using eq. $\left(1\right), \left(2\right), \left(3\right) \& \left(4\right)$

$a=\frac{e}{e^2-3e+13}$