Is this a measure on the sigma algebra of countable and cocountable subsets of R?

Let $A=\bigcup_{n=1}^{\infty}A_n$ with $A_n\in\Sigma$ disjoint.

If $A_n$ is countable for each $n$ then also $A$ is countable so that: $$\mu(A)=0=\sum_{n=1}^{\infty}\mu(A_n)$$

This was allready noted by yourself.

Now let it be that e.g. $A_1$ is cocountable.

Then also $A$ is cocountable. Secondly the disjointness of the sets $A_n$ implies that $A_n\subseteq A_1^c$ for $n=2,3,\dots$ so as subsets of a countable set these sets must be countable.

Then:$$\mu(A)=1=1+0+0+\cdots=\sum_{n=1}^{\infty}\mu(A_n)$$

It is obvious that this also works if another set (e.g. $A_{45}$) is taken to be cocountable.

Proved is now that $\mu$ is $\sigma$-additive.

Assume that there are two disjoint uncountable sets $A, B$ with countable complements. Then $\mathbb{R} = (A\cap B)^c = A^c \cup B^c$ would be the union of two countable sets, in contradiction to the uncountability of the real numbers.