We can be sure, that $$\prod\limits_{p}^{\infty}\frac{p}{p-1}\prod\limits_{p[4k+1]}^{\infty}\frac{p-1}{p+1}=\pi$$ and $$\prod\limits_{p}^{\infty}\frac{p}{p-1}\prod\limits_{p[6k+1]}^{\infty}\frac{p-1}{p+1}=\frac{2\pi}{\sqrt{3}}$$ so $$\frac{2}{\sqrt{3}}=\prod\limits_{p[4k+1]}^{\infty}\frac{p+1}{p-1}\prod\limits_{p[6k+1]}^{\infty}\frac{p-1}{p+1}=\left(\frac{5+1}{5-1}\right)\left(\frac{7-1}{7+1}\right)\left(\frac{17+1}{17-1}\right)\left(\frac{19-1}{19+1}\right)\cdots$$ For all $p[12k+1]$ that production is equal to $1$, so $$\frac{2}{\sqrt{3}}=\frac{3}{2}\cdot\frac{3}{4}\cdot\frac{9}{8}\cdot\frac{9}{10}\cdot\frac{15}{14}\cdot\frac{15}{16}\cdots$$ It might seem, that this result is equal to $$\prod\limits_{n=1}^{\infty}\left[\frac{(6n-3)^2}{(6n-2)(6n-4)}\right]=\frac{2}{\sqrt{3}}$$ but it actually is. This two series are similar, but different. There are some terms in second infinite production, which are absent in first: $$\frac{2}{\sqrt{3}}=\frac{3}{2}\cdot\frac{3}{4}\cdot\frac{9}{8}\cdot\frac{9}{10}\cdot\frac{15}{14}\cdot\frac{15}{16}\cdot \frac{21}{20}\cdot\frac{21}{22}\cdot\frac{27}{26}\cdot\color{red}{\frac{27}{28}}\cdot\color{red}{\frac{33}{32}}\cdot\frac{33}{34}\cdot\color{red}{\frac{39}{38}}\cdot\frac{39}{40}\cdot\frac{45}{44}\cdot\color{red}{\frac{45}{46}}\cdot\frac{51}{50}\cdot\frac{51}{52}\cdots$$ Part of them smaller than $1$, other part - bigger. Also there is no terms in first production, which are absent in second. So if we extract this (colored red) terms, their infinite production equal to $1$. Another words, we can say, that $$\prod\limits_{s[12k+5]}^{\infty}\frac{s+1}{s-1}\prod\limits_{s[12k+7]}^{\infty}\frac{s-1}{s+1}=1$$
Here $s[mk+n]$ - composite number of form $mk+n$. But why?
Why are they equal to one result?
If I made some mistakes, sorry for my English.
There is a flaw in your so part:
According to you, if $a_p$ and $b_p$ are two series: $$\prod a_p \, \prod b_p = \zeta \implies \prod a_p = \zeta \prod \frac1{b_p}$$
How is this true? For example, take $b_p= p^2$. Does the result follow?