This is not true. Indeed, suppose that $X_k=X_{s;k}=k+sZ_k$, where $s\downarrow0$ and the $Z_k$'s are any iid random variables r(r.v.'s).
To obtain a contradiction, 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 \begin{equation} \mu_s((-1/2,1/2))\to1 \tag{1} \end{equation} in probability (see details on (1) below). Therefore and because the r.v. $\mu_s((-1/2,1/2))$ has the Poisson distribution with parameter $\lambda(s)$, necessarily $\lambda(s)\to0$ and hence $\mu_s((-1/2,3/2))\to1$ in probability. However, similarly to (1) we have $\mu_s((-1/2,3/2))\to2$ in probability, a contradiction.
So, the random measure $\mu_s$ cannot be Poisson for all $s>0$.
Proof of (1): Note that $\mu_s(B)=\sum_{k\in\mathbb Z}1(Z_k\in\frac{B-k}s)$ and hence
\begin{equation}
1-\mu_s((-1/2,1/2))=s_1-s_2,
\end{equation}
where
\begin{equation}
s_1:=1-1\Big(Z_0\in\Big(\frac{-1/2}s,\frac{1/2}s\Big)\Big),
\end{equation}
and
\begin{equation}
s_2:=\sum_{k\in\mathbb Z\setminus\{0\}}1\Big(Z_k\in\Big(\frac{-1/2-k}s,\frac{1/2-k}s\Big)\Big).
\end{equation}
Next,
\begin{equation}
Es_1=1-P\Big(Z_0\in\Big(\frac{-1/2}s,\frac{1/2}s\Big)\Big)\to0
\end{equation}
and
\begin{equation}
Es_2=\sum_{k\in\mathbb Z\setminus\{0\}}P\Big(Z_0\in\Big(\frac{-1/2-k}s,\frac{1/2-k}s\Big)\Big)\le Es_1.
\end{equation}
SoTherefore and because $s_1,s_2\ge0$, we have
\begin{equation}
E|\mu_s((-1/2,1/2))-1|\le Es_1+Es_2\to0.
\end{equation}
So, by Markov's inequality, (1) follows.