Since $x$ is constant inside the integral, write

$f(x)=e^x+A+Bx{e}^x$, where $A={\int }_0^{1}yf(y)dy$ and $B={\int }_0^{1}f(y)dy$.

Compute $B$: $B={\int }_0^1(e^y+A+By{e}^y)dy=(e-1)+A+B\cdot 1$, since ${\int }_0^{1}y{e}^{y}dy=1$.

$B-B=e-1+A\Rightarrow A=1-e$.

(The value of $B$ itself is also needed for $f$, but $f(0)$ depends only on $A$:)

$f(0)=e^0+A+B\cdot 0\cdot e^0=1+A=1+(1-e)=2-e$.

$\therefore e+f(0)=e+(2-e)=2$.