Your sum actually equals $\frac{\pi\sqrt{3}}{2}$, so it's more like $\pi$ all over again, not $e$. To see this, note first that by definition $\mathrm{sgn}_2$ is multiplicative, hence $$ A=\sum_{n=1}^{+\infty}\frac{\mathrm{sgn}_2(n)}{n}=\prod_p\left(1-\frac{\mathrm{sgn}_2(p)}{p}\right)^{-1}=\left(1-\frac12\right)^{-1}\left(1-\frac13\right)^{-1}\prod_{p>3}\left(1-\frac{\mathrm{sgn}_1(p)}{p}\right)^{-1}=3B. $$$$ A=\sum_{n=1}^{+\infty}\frac{\mathrm{sgn}_2(n)}{n}=\prod_p\left(1-\frac{\mathrm{sgn}_2(p)}{p}\right)^{-1}= $$ $$ =\left(1-\frac12\right)^{-1}\left(1-\frac13\right)^{-1}\prod_{p>3}\left(1-\frac{\mathrm{sgn}_1(p)}{p}\right)^{-1}=3B. $$ Next, if you restrict $\mathrm{sgn_1}$ to $n$ with $(n,6)=1$ (and set it to $0$ elsewhere), you will get the quadratic character $\mod 6$, which is once again multiplicative. Therefore, $$ \prod_{p>3}\left(1-\frac{\mathrm{sgn}_1(p)}{p}\right)^{-1}=B=1-\sum_{n=1}^{+\infty}\left(\frac{1}{6n-1}-\frac{1}{6n+1}\right). $$ The last sum can be transformed into $$ B=1-\sum_{n=1}^{+\infty}\frac{2}{36n^2-1}=\frac{1}{36}\left(36+\sum_{n=1}^{+\infty}\frac{2}{\frac{1}{36}-n^2}\right). $$ From a well-known identity $$ \frac{\pi \cot\pi x}{x}=\frac{1}{x^2}+\sum_{n=1}^{+\infty}\frac{2}{x^2-n^2} $$ we deduce that $$ B=\frac{1}{36}\cdot6\pi\cot \frac{\pi}{6}=\pi\frac{\sqrt{3}}{6} $$ and $$ A=\frac{\pi\sqrt{3}}{2}. $$ One can also notice that $$ \frac{\pi\sqrt{3}}{2}-e\approx 0.0024, $$ so no, these two constants are not even that close.
As for the reference for more general series of this type, see H. Davenport "Multiplicative number theory", Chapter 6 "Dirichlet's class number formula" or Wikipedia page on the same formula