A powerful method to prove this result (and many similar ones) is to use the fact that a language is regular if and only if it is recognized by a finite monoid. A language $L$ of $A^*$ is recognized by a finite monoid $M$ if there is a surjective monoid morphism
$f:A^* \to M$ and a subset $P$ of $M$ such that $f^{-1}(P) = L$.
Let $\mathcal{P}(M)$ be the monoid of subsets of $M$, with product defined, for each $X, Y \in \mathcal{P}(M)$, by
$$
XY = \{xy \mid x \in X, y \in Y\}
$$
Let now $N$ be the commutative submonoid of $\mathcal{P}(M)$ generated by $f(A)$. I claim that $L_{1/3}$ is recognized by $M \times N$ and hence is regular.
Indeed, let $g: A^* \to M \times N$ be the monoid morphism defined, for each letter $a \in A$, by $g(a) = (f(a), f(A))$. Thus, for each word $u \in A^*$,
$$
g(u) = (f(u), f(A)^{|u|})
$$
Setting
$$
Q = \{ (m,R) \in M \times N \mid RmR \cap P \not= \emptyset \},
$$
one gets
\begin{align}
g^{-1}(Q) &= \{ u \in A^* \mid g(u) \in Q \} = \{ u \in A^* \mid f(A)^{|u|}f(u)f(A)^{|u|} \cap P \not= \emptyset \}\\
&= \{ u \in A^* \mid f(A^{|u|}uA^{|u|}) \cap P \not= \emptyset \}
= \{ u \in A^* \mid (A^{|u|}uA^{|u|}) \cap f^{-1}(P) \not= \emptyset \} \\
&= \{ u \in A^* \mid A^{|u|}uA^{|u|} \cap L\not= \emptyset \}\\
&= L_{1/3}
\end{align}
which proves the claim.