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 $\mathbb{N} \backslash (\mathbb{N} + 1) = \{1\}$, so $\{1\} \in \mathcal{A}$. By translation again, $\mu(\{n\}) = \mu(\{1\})$ for any $n \in \mathbb{N}$. Finally, abusing notation to call the extension $\mu$, for any set $A \subseteq \mathbb{N}$, we have $\mu(A) = \sum_{a \in A} \mu(\{a\}) = \mu(\{1\}) \sum_{a \in A} 1 = |A| \mu(\{1\})$. Therefore, any measure that is translation-invariant and countably additive must be the counting measure (up to scaling). Further, if this measure is finite (that is, if $\mu(\mathbb{N}) < \infty$, then it must be zero.