Using the Power Series expansion for the natural exponential,
\begin{align*}
\sum_{x=0}^{\infty} f(x) & = \sum_{x=0}^{\infty} \frac{(\lambda T)^x}{x!} e^{-\lambda T} \\
& = e^{-\lambda T} \sum_{x=0}^{\infty} \frac{(\lambda T)^x}{x!} \\
& = e^{-\lambda T} e^{\lambda T} \\
& = 1
\end{align*}