I'd like to know what I've done wrong when I tried to solve the functional equation below for any domain and codomain where $f$ exists and its derivate as well:
$f(x+c)=f(x)+ce^{x}$
$f(x+c)-f(x)=ce^{x}$
$\displaystyle\frac{f(x+c)-f(x)}{c}=e^{x}$
$lim_{c\rightarrow0}\displaystyle\frac{f(x+c)-f(x)}{c}=lim_{c\rightarrow0}e^{x}$
$df(x)/dx=e^{x}$
$f(x)=e^{x}$
but $e^{x+c}=(c+1)e^{x}$ doesn't hold, just take $c=1$ and $x=0$ then we have $e=2$ xD.
Assuming $f$ is a differentiable function such that for all $x,c\in\Bbb{R}$, $f(x+c)=f(x)+ce^x$, you've correctly shown that $f'(x)=e^x$ for all $x\in\Bbb{R}$. From here, it follows that there is come constant $A\in\Bbb{R}$ such that for all $x\in\Bbb{R}$, $f(x)=e^x+A$. So $f(x+c)=f(x)+ce^x$ now implies $e^{x+c}+A=e^x+A+ce^x$, which implies $e^ce^x=(1+c)e^x$ for all $c,x\in\Bbb{R}$. By choosing $x=0$, we then get $e^c=1+c$ for all $c\in\Bbb{R}$, which is absurd. The exponential cannot equal an affine function.
What this shows is that the original functional equation you started with has no solutions.