Here is a detailed solution to the modified -- substantially more challenging -- problem (see the OP's comments below the previous answer; in particular, it is stated there that this is not homework).
To find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation $ f(t) = \int_0^t {u e^u f(t - u)\,du} + e^t, \;\; t \geq 0, $ begin, as in the previous answer, by writing $ \hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1, $ where this time $\hat \varphi$ is the Laplace transform of the convolution $ (f*te^t )(t) = \int_0^t {ue^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)(t-u)e^{t - u}\,du}\bigg). $ By the convolution theorem, $ \hat \varphi(s) = \hat f(s) \frac{1}{{(s - 1)^2 }}. $ It is worth noting that the term $1/(s-1)^2$ can be derived as follows, recalling that an exponential random variable with density function $\lambda e^{-\lambda t}$, $t \geq 0$, has mean equal to $1/\lambda$ (here \lambda = s -1 > 0): $ \int_0^\infty {e^{ - st} te^t \,dt} = \frac{1}{{s - 1}}\int_0^\infty {t(s - 1)e^{ - (s - 1)t} \,dt} = \frac{1}{{(s - 1)^2 }}. $ Solving for $\hat f(s)$ (using the above expression for $\hat \varphi(s)$) gives $ \hat f(s) = \frac{{s - 1}}{{s^2 - 2s}} = \frac{{(s - 2) + 1}}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{s}\frac{1}{{s - 2}}. $ Assuming that $s > 2$, it follows by inversion (and the Convolution Theorem) that $ f(t) = 1 + (1 * e^{2t})(t),\;\; t \geq 0. $ (Indeed, note that $\int_0^\infty {e^{ - st} 1\,dt} = \frac{1}{s}$ and $\int_0^\infty {e^{ - st} e^{2t} \,dt} = \frac{1}{{s - 2}}$.) Finally, from $ (1 * e^{2t})(t) = \int_0^t {e^{2u} 1\,du} \,\bigg( = \int_0^t {1e^{2(t - u)} \,du} \bigg) , $ it follows that $ f(t) = 1 + \frac{{e^{2t} - 1}}{2} = \frac{{e^{2t} + 1}}{2},\;\; t \geq 0. $ Indeed, this $f$ satisfies the original equation; that is, as one can easily verify, it holds $ \frac{{e^{2t} + 1}}{2} = \int_0^t {ue^u \frac{{e^{2(t - u)} + 1}}{2}\,du} + e^t . $
EDIT (in response to the OP's comment below). While inverting $1/s$ gives $1$ and inverting $1/(s-2)$ gives $e^{2t}$, inverting $1/(s(s-2))$ does not give the product of $e^{2t}$ and $1$; rather, by the Convolution Theorem, it gives the convolution of $e^{2t}$ and $1$. Since $(1 * e^{2t})(t) = \frac{{e^{2t} - 1}}{2}\,( = \int_0^t {e^{2u} \,du} )$, $ f(t) = 1 + \frac{{e^{2t} - 1}}{2} = \frac{{e{}^{2t} + 1}}{2} $ (which was verified by substitution into the original equation). However, as the OP observed, the solution can be obtained more elementarily by splitting $\hat f(s)$ into partial fractions. Specifically, $ \hat f(s) = \frac{{s - 1}}{{s^2 - 2s}} = \frac{1}{{2s}} + \frac{1}{{2(s - 2)}}, $ from which it follows, by inversion, that $ f(t) = \frac{1}{2} + \frac{1}{2}e^{2t} = \frac{{e^{2t} + 1}}{2}. $