A classical inequality of Teicher (1955) asserts
Proposition (Teicher). Let $X \sim \mathrm{Pois}(\lambda)$. Then, $\mathbb P(X \leq \lambda) > e^{-1}$.
A modification of his argument will allow us to prove the following.
Proposition. For $\lambda \in (0,1)$, let $X_{n\lambda} \sim \mathrm{Pois}(n\lambda)$. Then the sequence $A_n := A_{n,n\lambda} = \mathbb P(X_{n\lambda} < n)$ is monotonically nondecreasing. In particular, $A_n \geq e^{-\lambda}$ for all $n$.
Proof. First, note that, by considering a Poisson process with rate 1, we have, $$ A_{n+1,\mu} = \sum_{x=0}^n \frac{e^{-\mu} \mu^x}{x!} = \int_\mu^\infty \frac{x^n e^{-x}}{n!} \\,\mathrm{d}x \\,, $$ for all $n$ and $\mu$. Now, $$ A_{n+1,n\lambda} - A_{n+1,(n+1)\lambda} = \int_{n\lambda}^{(n+1)\lambda} \frac{x^n e^{-x}}{n!} \\,\mathrm{d}x = \int_0^1 (\lambda(y+n))^n \frac{e^{-\lambda(y+n)}}{n!} \lambda \\,\mathrm{d}y \\,, $$ where the last equality follows from the substitution $y = (x-n\lambda)/\lambda$.
We can rewrite the integral as $$ \frac{e^{-\lambda n}(\lambda n)^n}{n!} \lambda \int_0^1 (1+y/n)^n e^{-\lambda y} \\,\mathrm{dy} \leq \frac{e^{-\lambda n}(\lambda n)^n}{n!} = A_{n+1,n\lambda} - A_{n,n\lambda} \\, $$ where the inequality follows from facts that $(1+y/n)^n < e^y$ and (upon integrating) $e^{1-\lambda} \leq \lambda^{-1}$, true for any $\lambda \in (0,1)$.
But, then $A_{n,n\lambda} \leq A_{n+1,(n+1)\lambda}$ which is what was to be shown. Since $A_1 := A_{1,\lambda} = e^{-\lambda}$, the second part of the proposition statement holds.