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