This cannot be true in general. E.g., suppose that $X_k=X_{s;k}\sim N(k,s^2)$, with $s\downarrow0$. Suppose that, for the random Borel measure $\mu_s$ over $\mathbb R$ defined by $\mu_s(B):=\sum_{k\in\mathbb Z}1(X_{s;k}\in B)$, the distribution of the random variable (r.v.) $\mu_s(B)$ is Poisson with parameter $\lambda(s)|B|$ for some $\lambda(s)>0$ and all Borel $B$, where $|B|$ is the Lebesgue measure of $B$. 

Note that $\mu_s((-1/2,1/2))\to1$ in probability. So, necessarily $\lambda(s)\to0$ and hence $\mu_s((-1/2,3/2))\to1$ in probability. However, in fact $\mu_s((-1/2,3/2))\to2$ in probability, a contradiction. 

So, the random measure $\mu_s$ cannot be Poisson for all $s>0$.