Highly grateful for your help/steers on the following question (at the end):

Take the infinite product:

$$\displaystyle T(s) = \prod _{n=2}^{\infty } \left( \dfrac{{n}^{s}} {{n}^{s}-1}\right)$$

for $\Re(s) > 1$ it is equal to:

$$\displaystyle \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) * \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1}\right)$$

I.e. the Euler-product (equal to $\zeta(s)$) multiplied by its composite "equivalent" ( excluding 1 since that is a bit of a strange composite anyway).

Why my interest? I wanted to learn more about the composite infinite product (and see if it had a 'zeta' like version). Soon became clear to me that the only way to learn more about this product, is to concentrate on $T(s)$ and then divide it by $\zeta(s)$.

I searched the web but there is hardly anything known about $T(s)$. F.i. Wolfram math only shows (formula 20) two different solutions (note: both need to be raised to $^{-1}$ to get $T(s)$ !) for odd and even integers and by reading through some arxiv math pre-prints the best I could find was a single, but still integer only formula that is:

$$\prod _{k=1}^{s-1}\Gamma \left( 2- {{\rm e}} ^{{\frac {2 i \pi k}{s}}} \right), ( \Re(s) > 1, s \in \mathbb{N})$$

I then decided to explore ways to extend the domain for $s$ and derived the following formula:

$$\displaystyle \ln \left( T\left( s \right) \right) = \ln \prod_{n=2}^{\infty } \left( \left( -1+{n}^{-s} \right) ^{-1} \right) = \sum_{n=2}^{\infty } \ln \left( \left( 1-{n}^{-s} \right) ^{-1} \right)$$

$$\displaystyle = \sum_{m=1}^\infty \sum_{n=2}^{\infty } \frac{1}{mn^{ms}} = \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n}$$

And this brings us to:

$$ T(s)={\rm e}^{\left( \displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$$

Yep, there's always a $\zeta(s)$ hiding around the corner somewhere...

So, let's see what the plot looks like for $s>0$ ($T(s)$ diverges for $s<0$).

$T(s)=\displaystyle {\rm e}^{\left( \displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)} \text{ blue}$

$\displaystyle \zeta(s) = \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) \text{purple}$

$\displaystyle \frac{T(s)}{\zeta(s)} = \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1} \right) \text{ brownish}$

For $s>1$ I could numerically solve the following equation:

$$\displaystyle \prod _{primes}^{\infty } \left( \dfrac{{p}^{s}} {{p}^{s}-1}\right) = \prod _{composites}^{\infty } \left( \dfrac{{c}^{s}} {{c}^{s}-1} \right)$$

giving this interesting number $s = 1.397737620...$ (there is only one for $\Re(s) > 1$ )

I obviously took a deep dive with this number on Google and Plouffe's inverter, but have not found anything 'beautiful' or related to other constants as yet.

Then the domain $0 < s < 1$. It is easy to see in the graph that $T(s)$, and therefore also $\dfrac{T(s)}{\zeta(s)}$, have 'trivial' poles for $s= \dfrac{1}{k}, k \in \mathbb{N}$ that are induced by the fact that for each $s= \dfrac{1}{k}$ there always is a $n s = 1$ that makes at least one term in the infinite sum equal to the pole $\zeta(1)$ (hence the whole sum turns into a pole).

But I'm actually mostly intrigued by what happens under the x-axis and especially where:

$$\zeta(s) = \dfrac{T(s)}{\zeta(s)}$$ or

$$|\zeta(s)| = {\rm e}^{\displaystyle \left(\frac12 \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$$.

If I have done my analysis correctly, this result would imply that there are an infinite number of values for $0 < s < 1$, where the (analytically continued) infinite products of primes and composites are equal (since $\zeta(s)$ remains negative between $0 < s < 1$ and there are an infinite number of poles separating the intersection points). And that would imply/reveal an infinite amount of tiny bits of information about how the primes 'grow like weed between the composites'.

Of course I checked $T(s)$ also for $s \in \mathbb{C}$, however, any graph I've produced sofar for $s=a+bi$ of $T(s)={\rm e}^{\left(\displaystyle \sum_{n=1}^\infty \frac{\zeta(n s)-1}{n} \right)}$ did not reveal any non-trivial zeroes (nope, not even at $a=\frac12$...), although the curves do seem to be trending towards a large number of very chaotically distributed zeroes when $a \rightarrow 0$.

So, apologies for the relatively long intro to my question:

Since $\zeta(s)$ has been analytically continued throughout the entire complex domain, is it allowed to also analytically continue the division of $\dfrac{T(s)}{\zeta(s)}$ into the domain $s<1$? Or do the nominator and denominator each require an individual continuation and does the concept of division get 'lost in continuation'?