I am looking for a function $f(x)$ such that \begin{align} E[f(X)]=\lambda \log (\lambda) \quad \text{ for all } \lambda \ge 0 \tag{$*$} \end{align} where $X$ is a Poisson random varaible with parameter $\lambda$. Note, we are looking for a function idenpendent of $\lambda$.
Here are some thoughts: \begin{align} \lambda \log (\lambda) = E[f(X)]= \sum_{k=0}^\infty f(k) \frac{\lambda^k e^{-\lambda}}{k!} \end{align} Therefore, we have that \begin{align} e^{\lambda} \lambda \log (\lambda) = \sum_{k=0}^\infty f(k) \frac{\lambda^k }{k!}= \sum_{k=0}^\infty a_k \frac{\lambda^k }{k!} \tag{$**$} \end{align} where in the last step I defined $f(k)=a_k$.
Now, this is the point where I am a bit stuck.
It doesn't look like the expression in $(**)$ can hold as I don't think the function $g(x)=e^x x \log(x)$ can be written as a Maclaurin series. These lets me to conclude that there is no function $f(x)$ such that $(*)$ holds.
To be precise, we want to find $f:\mathbb{N}_0\to\mathbb{C}$ such that $ E\left[|f(X)|\right]<\infty, $ and $$ E[f(X)]=\lambda\log\lambda,\quad\forall \lambda\ge 0. $$
Now, we can write $$ \sum_{k=0}^\infty f(k)\frac{\lambda^k e^{-\lambda}}{k!}=\lambda\log\lambda, $$or equivalently $$ \sum_{k=0}^\infty f(k)\frac{\lambda^k}{k!}=\lambda e^\lambda\log\lambda . $$ The LHS is by the assumption an absolutely convergent power series for all $\lambda\ge 0$. Then this implies that it can be extended uniquely to some entire function $F(\lambda)$. Since $F(0) = \lim\limits_{\lambda\to 0^+}\lambda e^\lambda \log \lambda =0$, $G(\lambda)=\frac{F(\lambda)} {\lambda}$ also defines an entire function. Now, we have $$ e^{-\lambda}G(\lambda) =\log\lambda $$ for all $\lambda>0$, but $\lim\limits_{\lambda\to 0^+}e^{-\lambda}G(\lambda)=G(0)\neq \lim\limits_{\lambda\to 0^+}\log\lambda=-\infty.$ This leads to a contradiction. Therefore, there does not exist $f$ such that $E[|f(X)|]<\infty$ and $E[f(X)]=\lambda \log \lambda$ for all $\lambda\ge 0$.