Weird summation

First, we know $\sum_{n=1}^\infty \dfrac{F(n)}{n^s}=\zeta(s)\sum_{p=prime}\dfrac{F(p)}{p^s-1}$ For F being a completely additive function F. proved here .

Using this we have $\sum_{n=1}^\infty \dfrac{\omega_{-s}(n)}{n^s}=\zeta(s)\sum_{p=prime}\dfrac{p^{-s}}{p^s-1}\\=\zeta(s)\sum_{p=prime} \left(\dfrac{1}{p^s-1}-\dfrac{1}{p^s}\right) =\zeta(s)\sum_{p=prime} \dfrac{1}{p^s-1} -\zeta(s)P(s)$ By the form we used, we deduce $\zeta(s)\sum_{p=prime} \dfrac{1}{p^s-1}=\sum_{n=1}^\infty \dfrac{\Omega_0 (n)}{n^s}$ and we can work out $\zeta(s)P(s)=\sum_{n=1}^\infty \dfrac{\omega (n)}{n^s}$ So, $f(n)=(\Omega_0-\omega)(n)$ The result follows.