Note that $d(A,B) = \| 1_A-1_B\|_1$, so we can use properties of $L^1(\mathbb{R})$ to show completeness.
As Michael noted in the comments above, one needs to be careful with infinities.
Suppose we have a $d$-Cauchy sequence of sets $A_n$. Without loss of generality
we can presume that
$d(A_n,A_1) < 1$ for all $n$.
Let $f_n = 1_{A_n} - 1_{A_1}$. We have $f_n \in L^1(\mathbb{R})$ for all $n$
and $f_n$ is Cauchy. Since $L^1(\mathbb{R})$ is complete, we know that there
is some $f \in L^1(\mathbb{R})$ such that $f_n \to f$ (in the $\|\cdot\|_1$ norm, of course).
The only issue remaining is to 'extract' a set $A$ such that $f(x) = 1_A(x)- 1_{A_1}$ ae. $[\mu]$.
A standard result (see https://math.stackexchange.com/a/716328/27978, for example) is that if $f_n \to f$ in $L^1(\mathbb{R})$, then there is a
subsequence such that $f_{n_k}(x) \to f(x)$ ae. $[\mu]$. We note that
$f_n(x) \in \{-1,0,+1\}$ for all $n$, hence we have
$f(x) \in \{-1,0,+1\}$ ae. $[\mu]$. Let $N= f^{-1}(\{-1\})$, $Z= f^{-1}(\{0\})$ and $P= f^{-1}(\{+1\})$ and let $A = (Z \cap A_1) \cup P$, then
it is easy to check that $f(x) = 1_A(x) - 1_{A_1}(x)$ ae. $[\mu]$ and
$d(A,A_n) \to 0$. (Measurability follows from the fact that $f \in L^1(\mathbb{R})$.)