Given $f\left(x\right)={\int }_0^x{e}^{t}f\left(t\right)dt+e^x\Rightarrow f\left(0\right)=1$

Differentiating with respect to $x$ we get, 

$f^{\text{'}}\left(x\right)=e^{x}f\left(x\right)+e^x$

$f^{\text{'}}\left(x\right)=e^x\left(f\left(x\right)+1\right)$

$\frac{f^{\text{'}}\left(x\right)}{f\left(x\right)+1}=e^x$

Integrate both the sides we get, 

${\int }_0^x\frac{f^{\text{'}}\left(x\right)}{f\left(x\right)+1}dx={\int }_0^x{e}^{x}dx$

${\overline{)\ln \left(f\left(x\right)+1\right)}}_0^x{\overline{)=e^x}}_0^x$

$\ln \left(f\left(x\right)+1\right)-\ln \left(f\left(0\right)+1\right)=e^x-1$

$\ln \left(\frac{f\left(x\right)+1}{2}\right)=e^x-1$   {as $f\left(0\right)=1$}

$f\left(x\right)=2e^{\left(e^x-1\right)}-1$