A zero-truncated Poisson distribution has the probability mass function $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k! (1 - e^{-\lambda})}$ for $k \in \{1, 2, 3, \dots\}$, where $\lambda > 0$. If the expected value of this distribution is exactly $2$, determine the value of $\lambda + 2 e^{-\lambda}$.