You recently asked a similar question for modulus $4$ on math.stackexchange. I just used the exact same technique.

For $i\in\{1,5\}$, define $f_i:\mathbb{Z}_{\ge 1}\to \mathbb{C}^\times$ by $f_i(n)=(-1)^{\Omega_i(n)}$, then $f_i$ is completely multiplicative and
$$L(s,f_i)=\left(1-2^{-s}\right)^{-1}\left(1-3^{-s}\right)^{-1}\prod_{p\equiv i\pmod 6}\left(1+p^{-s}\right)^{-1}\prod_{p\equiv -i\pmod 6}\left(1-p^{-s}\right)^{-1}.$$
Hence,
$$L(s,f_1)L(s,f_5)=\left(1-2^{-s}\right)^{-2}\left(1-3^{-s}\right)^{-2}\prod_{p\ge 5}\left(1-p^{-2s}\right)^{-1}=\frac{2^{s}+1}{2^s-1}\cdot\frac{3^s+1}{3^s-1}\cdot \zeta(2s).$$
Therefore, if both $L(1,f_1)$ and $L(1,f_5)$ converge, their product is $6\zeta(2)=\pi^2$. Now, let $\chi$ be the non-trivial Dirichlet character modulo $6$, then
$$L(1,\chi)=\prod_{ p\equiv 1\pmod 5}(1-p^{-1})^{-1}\prod_{p\equiv 5\pmod{6}}\left(1+p^{-1}\right)^{-1}=\frac13L(1,f_5).$$
Therefore, it suffices to show that $L(1,\chi)=\frac{\pi}{2\sqrt{3}}.$

Let $\chi_2$ be the non-trivial Dirichlet character modulo $3$, then $\chi(p)=\chi_2(p)$ for all primes $p\neq 2$ and
$$L(1,\chi)=\frac{L(1,\chi_2)}{(1+2^{-1})^{-1}}=\frac32 \cdot \frac{\pi}{3\sqrt 3}=\frac{\pi}{2\sqrt{3}}$$
Where we use this answer.

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