Let $x_1, \ldots, x_n$ be observations of random vars $X_1, \ldots, X_n$ with $X_i \sim \mathcal{P}(\lambda)$ a Poisson distribution. It is straigthforward to prove that the sample mean $\bar{X} := 1/n \sum x_i$ is the MLE of $\lambda$. But what about the standard deviation of this estimator? My approach has been the following.
We know the sum of $n$ i.i.d. Poisson variables follows the distribution $\mathcal{P}(n\lambda)$. Let $\Sigma$ denote this sum so that $\overline{X} = \frac{1}{n}\Sigma$. Then
\begin{align*} P(\overline{X} \leq x) &= P(\frac{1}{n} \Sigma \leq x) \\ &=P(\Sigma \leq xn) \\ &=\int_0^{xn} \frac{\lambda^{xn} e^{-\lambda}}{(xn)!} ~ dx \end{align*}
Letting $xn = u$ s.t. $du = n dx$ we have
\begin{align*} P(\Sigma \leq u) &= \frac{1}{n}\int_0^{u} \frac{\lambda^u e^{-\lambda}}{u!} ~ du \end{align*}
Since $\Sigma \sim \mathcal{P}(\lambda n)$, it follows that $\overline{X} \sim (\lambda)$. Then the variance of $\overline{X}$ is $\lambda$ and the standard deviation of the estimator is $\sqrt{\lambda}$.
Is this proof correct? I am concerned about whether the $u$ substitution performed should alter the conclusion that $\bar{X} \sim \mathcal{P}(\lambda)$.
You are correct that the sample total is Poisson distributed with intensity $n \lambda$, i.e. $$X_1 + \cdots + X_n \sim \operatorname{Poisson}(n \lambda),$$ but your conclusion regarding the standard deviation of the sample mean is incorrect for the reason that $$\operatorname{Var}[a X] = a^{\color{red}{2}} \operatorname{Var}[X]$$ for a scalar $a$. So while it is true that $$\operatorname{Var}[X_1 + \cdots + X_n] = n \lambda,$$ the variance of the sample mean is actually $$\operatorname{Var}[\bar X] = \operatorname{Var}\left[\frac{X_1 + \cdots + X_n}{n}\right] = \color{red}{\frac{1}{n^2}} \operatorname{Var}[X_1 + \cdots + X_n] = \frac{n \lambda}{n^2} = \frac{\lambda}{n}.$$ Consequently, the standard deviation of the sample mean is $$\sigma_{\bar X} = \sqrt{\operatorname{Var}[\bar X]} = \sqrt{\frac{\lambda}{n}}.$$ Note that this result also makes intuitive sense: the standard deviation of a small number of observations will be larger than the standard deviation of a huge number of observations--in other words, the variation of the sample mean should decrease as the sample size increases. While this property is not always satisfied for any random variable, it is true in the Poisson case.