Note if it was a turned into a (countably additive) measure, then $\mu(\{n\}) = 0$ for all $n \in \mathbb{N}$, which would mean $\mu(\mathbb{N}) = 0$, which I don't think you want. However, one wonders if you can turn it into a finitely additive measure. And indeed, finitely additive, shift invariant measures on $\mathbb{N}$ do exist, and they equal the natural density on subsets where the natural density exists. One way to construct this measure is to actually create it out of natural density, by viewing natural density as a linear functional on a certain subspace of $[0,1]^\mathbb{N}$ (where $A \subseteq \mathbb{N} \sim (1_A(n))_{n=1}^\infty$), and then extending it to a linear functional on all of $[0,1]^\mathbb{N}$ via Hahn-Banach.