My attempt:
We must show that $\forall \epsilon>0 \space \exists\delta>0 \space 0<|x|<\delta \Longrightarrow |f(x) - L| < \epsilon$
Rough work:
$|f(x) - L| = |e^x - 1| < \epsilon.$ Therfore, $-\epsilon < e^x - 1 < \epsilon => \ln(1-\epsilon) < x < \ln(1+\epsilon)$ Hence, $|x| < \ln(1+\epsilon)$.
Proof: Let $\epsilon > 0$ be given and $\delta = \ln(1+\epsilon)$. We know that $0<|x| < \delta$, and that $|f(x) - L| = |e^x - 1|$. Therfore, $|e^x - 1|< |e^{\delta} - 1| = |e^{\ln(1+\epsilon)} - 1| = |1 + \epsilon - 1| = \epsilon$