I know the answer must be $\frac{1}{2}$, because I have done the numerical simulation and as the parameter of the poisson distribution increases, the quantity gets closer to 0.5. Can we show this analytically?
Edit: For whoever comes across this in the future, you can solve this by expanding $X \log X$ into its Taylor series around $\lambda$. After some algebra, you can show that $E[X \log X] - \lambda \log \lambda = \frac{1}{2} + \frac{1}{12\lambda} + \frac{1}{12 \lambda^2} + O(\frac{1}{\lambda^3})$
Using the equation above we now have: $\lim_{\lambda \rightarrow \infty} \big[ E[X \log X] - \lambda \log \lambda \big] = \frac{1}{2}$