No closed form expression in terms of elementary functions, but if you are satisfied with a special functions (incomplete Gamma function $\Gamma$ and exponential integral Ei), then
$$\displaystyle\sum_{k=1}^\infty \frac{\lambda^k e^{-\lambda}}{k!}\cdot[1-(1-x)^k]\cdot \frac{1}{k}=\text{Ei}\left(\frac{\lambda}{e}\right)+\log (x-1)+\Gamma \left(0,\frac{\lambda (x-1)}{e}\right)$$