I would have a proof of the following fact; but it's a bit clunky, and am wondering if one can get a more elegant one (and/or improve the constants). I couldn't find this anywhere, and searching properties of the Skellam distribution didn't help much either.

Let $X\sim\operatorname{Poi}(\lambda)$ and $Y\sim\operatorname{Poi}(\mu)$ be two independent random variables. Then $$ \max(|\lambda-\mu|, \frac{1}{40}\min(\sqrt{\lambda+\mu}, \lambda+\mu)) \leq \mathbb E[|X-Y|] \leq\min(|\lambda-\mu|+\sqrt{\lambda+\mu}, \lambda+\mu) $$

*Here is the current proof I came up with:*

The upper bound follows from Cauchy–Schwarz, as $\mathbb E[(X-Y)^2] = (\lambda-\mu)^2 + \lambda+\mu$; and by the triangle inequality, as $\mathbb{E}[|X-Y|]\leq \mathbb{E}[X+Y]$. The lower bound $|\lambda-\mu|$ is a direct consequence of Jensen's inequality; we now proceed to establish the second term.

Consider first the case $\lambda+\mu\geq 1$. We will use Paley–Zygmund, noting that for any $c\in(0,1)$ we have $$ \mathbb{E}[|X-Y|] \geq c\sqrt{\mu+\lambda}\cdot\mathbb{P}\{|X-Y|\geq c\sqrt{\mu+\lambda} \} $$ and $$ \begin{align*} \mathbb{P}\{|X-Y|&\geq c\sqrt{\mu+\lambda}\} \\ &= \mathbb{P}\{(X-Y)^2\geq c^2(\mu+\lambda)\} \\ &\geq \mathbb{P}\{(X-Y)^2\geq c^2\mathbb{E}[(X-Y)^2]\}\\ &\geq (1-c^2)^2\frac{\mathbb{E}[(X-Y)^2]^2}{\mathbb{E}[(X-Y)^4]}\\ &= \frac{ (1-c^2)^2(\lambda+\mu+(\lambda-\mu)^2)^2}{\lambda+\mu+6(\lambda^2+\mu^2)+6(\lambda-\mu)^2(\lambda+\mu)+(\lambda-\mu)^2+(\lambda-\mu)^4}\\ &\overset{(\dagger)}{\geq} (1-c^2)^2 \frac{(\lambda+\mu)^2+(\lambda-\mu)^4}{(\lambda+\mu)+10(\lambda+\mu)^2+3(\lambda-\mu)^4}\\ &\overset{(\ddagger)}{\geq} \frac{(1-c^2)^2}{11} \end{align*} $$ where $(\dagger)$ is using the AM-GM inequality and $(a-b)^2\leq (a+b)^2$; and $(\ddagger)$ relies on $\lambda+\mu \leq (\lambda+\mu)^2$, which holds since $\lambda+\mu\geq 1$. Taking $c=1/\sqrt{5}$ concludes the proof of this case, as then $\frac{c(1-c^2)^2}{11} > \frac{1}{40}$.

If $\lambda+\mu \leq 1$, we will use properties of the modified Bessel function of the first kind $I_0$ (namely, that it is nondecreasing on $[0,\infty)$, with $e^{-x}I_0(x) \geq 1-\frac{x}{2}$ for $x\in[0,1]$) to conclude, as $$ \begin{align} \mathbb{E}[|X-Y|] &\geq 1 - \mathbb{P}\{|X-Y|=0\} \\ &= 1- e^{-(\lambda+\mu)} I_0(2\sqrt{\lambda\mu}) \\ &\geq 1- e^{-(\lambda+\mu)} I_0(\lambda+\mu) \geq \frac{\lambda+\mu}{2} \end{align}$$ where the first inequality is the AM-GM inequality.