I hope to find a non-trivial probability measure $\mu:\mathcal B\subset\mathcal P(\mathbb Z)\to[0, 1]$ such that $\mu(A) = \mu(c+A),~\forall A\in\mathcal B,~\forall c\in\mathbb Z$.
Of course, no finite subset of $\mathbb Z$ could be $\mu-$measurable (as discussed here).
Furthermore, if $A\in \mathcal B$ has a least element $a\in A$ , then $\displaystyle S = \{a+1,a+2,\dots\} = \bigcup_{i\ge1} i+A$ is measurable, therefore, so is $R = -1+S^C = \{\dots, a-2, a-1\}$ and $(S\cup R)^C = \{a\}$. Contradiction!
The same reasoning shows no set $A\in\mathcal B$ can have a maximum.
In fact, all such measures are derived from counting measures on $\mathbb{Z} / n\mathbb{Z}$. We say a set $A$ is $n$-periodic if $A = A + n$.
To see why, let $(\mathbb{Z}, \Sigma, \mu)$ be a translational invariant measure. Let $A$ be the largest subset of $\Sigma$ that doesn't contain $0$. As $\bar{A}$ is not finite, there exists some $n \neq 0$ not lying in $A$. Take such an $n$ with the smallest absolute value. Now note that $A - n$ does not contain $0$. So we have $$A - n \subset A.$$ As $-n \in A - n$, we have $-n \in A$. So by the same reasoning, $A + n \subset A$, thus $A \subset A - n$ and so $A = A - n$. So $A$ is $n$-periodic. By our choice of $n$, we can assume that $n > 0$, and conclude that $$A = \mathbb{Z} \backslash n\mathbb{Z}.$$ Using translational invariance, we now conclude that $i + \mathbb{Z} \backslash n \mathbb{Z} \in \Sigma$ for each $i \in [n]$. We can easily verify that all $n$-periodic sets lies in $\Sigma$.
To see these are all the sets, let $B$ be any element of $\Sigma$. If $i \notin B$, then $B - i$ does not contain $0$, so $B - i \subset A$. Therefore, if $i \notin B$, then $i + kn \notin B$ for any $k \in \mathbb{Z}$. This implies that $B$ is $n$-periodic, as desired. So $\Sigma$ consists exactly of $n$ periodic sets.
Now it is straightforward to verify that the only possible $\mu$ on $\Sigma$ is $$\mu(X) = c |X \cap [n]|$$ for some constant $c > 0$, as desired. (Indeed, one can take $c = \mu(\{kn: k \in \mathbb{Z}\})$, and use translational invariance.)