This can be proved similarly as Alexander Kalmynin's method .
Let, the sum be $S$, then we can make the following identity because $\text{sgn}_1$ of $2,3$ is defined to be $1$. So, $\text{sgn}_1(ak)=\text{sgn}_1(k), a=2,3,6$. Also, from the definition of $\text{sgn}_2$ we can see, $\text{sgn}_2(n)=1,-1$ respectively for $n \equiv 1,-1 (\text{modulo} 4)$.
As, $\text{sgn}_1(ak)=\text{sgn}_1(k), a=2,3$, we can separate them out from $S$ in the form of $\frac{S}{2}+\frac{S}{3}$. To prevent double counting of the $6$s multiples, we subtract $\frac{S}{6}$. Then, what is left is all odd numbers which aren't divisible by $3$. So, they can be classified as, $12k+\sigma , \sigma=1,-1,5,-5$ and also these are of the form $4k±1$.
The identity : $$S=\frac{S}{2}+\frac{S}{3}-\frac{S}{6}+\left(\sum_{n\geq 0}\frac{1}{12n+1}-\sum_{n\geq 1}\frac{1}{12n-1}\right)+\left(\sum_{n\geq 0}\frac{1}{12n+5}-\sum_{n\geq 1}\frac{1}{12n-5}\right)$$
This gives $\frac{S}{3}=(1+\frac{1}{5})+\frac{1}{12^2}\left(\sum_{n=1}^{\infty} \frac{2}{(\frac{1}{12})^2-n^2}+\sum_{n=1}^{\infty} \frac{ 10}{(\frac{5}{12})^2-n^2}\right)$
or, $\frac{S}{3}=\frac{\pi}{12}\left(\text{cot}(\frac{\pi}{12})+\text{cot}(\frac{5\pi}{12})\right)$
This gives $S=\pi$