The problem originates from the limit distribution of $\frac{N-\lambda}{\sqrt{\lambda}}$ as $\lambda \to \infty$, where $ N \sim \text{Po}(\lambda)$.
Expanding $\lambda (e^{\frac{it}{\sqrt{\lambda}}} -1)$ to the second order yields the terms
$$\lambda (e^{\frac{it}{\sqrt{\lambda}}} -1) = \lambda(1 + \frac{it}{\sqrt{\lambda}} - \frac{t^2}{2\lambda}+\mathcal{O}(\alpha(t)) -1)$$
The question is, what is the term $\alpha(t)$ and why does $\mathcal{O}(\alpha(t)) \to 0$ as $\lambda \to \infty$, for all $t$?
To follow up on the original question,
$$\varphi_{\frac{N-\lambda}{\sqrt{\lambda}}}(t) \to e^{-t^2/2} \ \text{as} \ \lambda \to \infty, $$
hence $\frac{N-\lambda}{\sqrt{\lambda}} \sim N(0,1).$