As I studied, the distribution of the sum of two independent random Poisson distributed variables is also Poisson distributed and we can take the sum of the parameters to get the $\lambda$ parameter of the sum. We also learned that the Poisson distribution is infinitely divisible, but then I don't really understand the following “counter(?)-example”...
Let's assume $\xi\sim Poi\left(\lambda\right)$, where $\lambda=1$. If we decompose $\xi$ as the sum of $\xi_{1}$ and $\xi_{2}$, where $\xi_{1}$ and $\xi_{2}$ are independent variables with the same Poisson distribution, then their parameters must be $\lambda/2=1/2$. Let's assume $\xi\overset{\circ}{=}\xi_{1}+\xi_{2}=1$. It indicates that $\left\{ \xi_{1}=1\;\text{and}\;\xi_{2}=0\right\}$ or $\left\{ \xi_{1}=0\;\text{and}\;\xi_{2}=1\right\}$. As a result, $$\frac{\left(\frac{1}{2}\right)^{1}}{1!}e^{-\frac{1}{2}}=\mathbf{P}\left(\xi_{1}=1\right)\geq\mathbf{P}\left(\xi=1\right)=\frac{1^{1}}{1!}e^{-1},$$ which inequality is not true.
So I'm a little bit confused.
That does not make sense to me. In fact $P\left(\xi_{1}=1\right)=\frac12e^{-1/2} \approx 0.303$ which is less than $P\left(\xi=1\right)=e^{-1} \approx 0.368$. But it is not obvious why you are comparing them.
If you want to do this then something like:
$P(\xi_{1}+\xi_{2}=1) =P(\xi_{1}=1 \text{ and }\xi_{2}=0) + P(\xi_{1}=0 \text{ and }\xi_{2}=1)$
$= \frac{\left(\frac{1}{2}\right)^{1}}{1!}e^{-\frac{1}{2}}\frac{\left(\frac{1}{2}\right)^{0}}{0!}e^{-\frac{1}{2}} + \frac{\left(\frac{1}{2}\right)^{0}}{0!}e^{-\frac{1}{2}}\frac{\left(\frac{1}{2}\right)^{1}}{1!}e^{-\frac{1}{2}}$
$= \frac{1}2e^{-\frac{1}{2}-\frac{1}{2}}+ \frac{1}2e^{-\frac{1}{2}-\frac{1}{2}} = e^{-1} $
$= \frac{1^{1}}{1!}e^{-1} =P(\xi=1)$ as hoped for