For $|\frac{x}{2\pi}|<1$ using Möbius inversion for the Taylor series of the logarithm one deduces $$\frac{x}{2\pi}=-\sum_{n=1}^\infty\frac{\mu(n)}{n}\log\left(1-\frac{x^n}{(2\pi)^n}\right)$$ thus $$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)}{kn}\frac{x^{nk}}{(2\pi)^{nk}}=\frac{1}{2\pi}(e^x-1)\sum_{k=0}^\infty\frac{B_k x^k}{k!},$$ where we've used the series expansion $\frac{x}{e^x-1}$ in terms of Bernoulli numbers.
Using Cauchy products in RHS I believe that it's possible deduce the following
Claim 1. For $|x|<2\pi$ $$\frac{1}{2\pi}(e^x-1)\sum_{k=0}^\infty\frac{B_k x^k}{k!}=\frac{x}{2\pi}+\frac{1}{2\pi}\sum_{n=2}^\infty\left(\frac{1}{n!}+\sum_{k=1}^{n-1}\frac{n!}{\binom {n} {k}}B_{n-k}\right)x^n.$$
Equating RHS of previous claim with LHS of our first deduction one has
(Wrong) Claim 2. For every composite integer $n\geq 2$ $$\sum_{\substack{a,b\geq 1}\\a\cdot b=n}\frac{\mu\left(\frac{n}{b}\right)}{n(2\pi)^n}=\frac{1}{2\pi}\left(\frac{1}{n!}+\sum_{k=1}^{n-1}\frac{n!}{\binom {n} {k}}B_{n-k}\right).$$
I am saying that this Claim 2 is wrong, the corresponding (correct) claim for prime numbers is:
If $n$ is prime then $$\frac{1}{n!}+\sum_{k=1}^{n-1}\frac{n!}{\binom {n} {k}}B_{n-k}=0.$$
Question. I would like to know the right claim for composite numbers. Many thanks.