There is no such measure.
Assume there is. Then we must have that $\mathbb{N} \in \mathcal{A}$. We can translate by $1$ to get $\mathbb{N} + 1 \in \mathcal{A}$. But $\mathcal{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. Moreover, $\mu(\mathbb{N} + 1) = \mu(\mathbb{N})$, so $\mu(\{1\}) = 0$. By translation again, $\mu(\{n\}) = 0$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \sum_{a \in A} 0 = 0$. Therefore, any measure that is translation-invariant and countably additive must be zero.