**Added:** In light of everything said, the easiest approach seems to be to just take $f_\epsilon(x) = \eta(x/\epsilon) e^{x-1}$, where $\eta$ is a smooth function with $\eta(0) = 0$ and $\eta(x) = 1$ for $x\geq 1$ (so that $\int_0^1 \eta'= 1$). The integral is then $e^{-1}\int_0^\epsilon \epsilon^{-1}\eta'(x/\epsilon)e^x\,dx = e^{-1}\int_0^1 \eta'(x) e^{\epsilon x}\,dx$, which tends to $e^{-1}$ as $\epsilon\to0$. I blame the desire to write things down in terms of elementary functions for the over-complication below.

Here's an idea for a sequence of minimizers. (As the comment points out your proof shows there's no actual minimizer.) Take $f_\alpha(x) = x^\alpha e^{x-1}$ and let $\alpha \to 0$; the point is to make the function look more and more like an exponential as the parameter is varied. Note that $f_\alpha'(x) - f_\alpha(x) = \alpha x^{\alpha - 1} e^{x-1}$, and making the substitution $y = x^\alpha$ in the integral should show that the integrals tend to $e^{-1}$ as $\alpha \to 0$.

**Edit:** John pointed out that $f_\alpha$ isn't differentiable at $0$. To fix this, take some smooth approximation to $f_\alpha$ instead, where the approximation becomes better (sufficiently fast) as the parameter tends to $0$. To be explicit, you could use $g_\alpha = \eta_\alpha f_\alpha$, where $\eta_\alpha$ is smooth and increasing with $\eta_\alpha(0) = 0$ and $\eta_\alpha(\epsilon) = 1$ for some small $\epsilon$ (depending on $\alpha$). For the approximation to work you'll just need to make sure that $\int_0^\epsilon \eta_\alpha'(x) x^\alpha \to 0$ with $\alpha$. You can arrange for $\eta_\alpha'\lesssim 1/\epsilon$, and then the integral will be $\lesssim\epsilon^\alpha$, so, for instance, $\epsilon = \epsilon(\alpha) = \alpha^{1/\alpha}$ should do.