Higher derivatives of weighted zeta function and continuation to $s=1$

Is it possible to analytically continue to weighted reciprocal zeta function $frac{1}{zeta_{a}(s)}$ to the point $s=1$ by bounding the quantity
begin{equation}
left|frac{d^n}{dx^n}prod_{kin S}left(1-frac{1}{k^x}right)right|tag{1}
end{equation}
for real $x>1$ and sets $S$ containinnng real numbers greater or equal to $2$?

Here the reciprocal weighted zeta function is defined by

begin{equation}
frac{1}{zeta_{a}(s)}=mathbb{E}_{ninmathbb{N}}left[a_nprod_{p|n}left(1-frac{1}{p^{s-1}}right)right]
end{equation}

where $a={a_1,a_2,a_3…}$ is a bounded sequence of real numbers. The motivation behind this definition is that using standard methods one can show that for any complex $s$ with $Re(s)>1$

begin{equation}
frac{1}{zeta(s)}=mathbb{E}_{ninmathbb{N}}left[prod_{p|n}left(1-frac{1}{p^{s-1}}right)right]tag{2}
end{equation}

and so by weighting (2) according to some set $a$ we obtain our definition.

I wish to show that $frac{1}{zeta_{a}(s)}$ has an analytic continuation to $s=1$, and to do this we can construct the power series of $frac{1}{zeta_{a}(s)}$ about $s=1+epsilon$ some small $epsilon>0$ and show that the radius of convergence of this power series is greater than $epsilon$, thus granting a continuation. To construct this power series we see that

begin{align*}
left|frac{d^n}{ds^n}frac{1}{zeta_{a}(s)}right|&=left|frac{d^n}{ds^n}mathbb{E}_{ninmathbb{N}}left[a_nprod_{p|n}left(1-frac{1}{p^{s-1}}right)right]right|\
&leqmathbb{E}_{ninmathbb{N}}left[left|a_nfrac{d^n}{ds^n}prod_{p|n}left(1-frac{1}{p^{s-1}}right)right|right]\
&leq Mmathbb{E}_{ninmathbb{N}}left[left|frac{d^n}{ds^n}prod_{p|n}left(1-frac{1}{p^{s-1}}right)right|right]
end{align*}

where $M=sup|a|$. I imagine that (1) has nice bounds, from which we could show that our power series has a suitable radius of convergence thus granting a solution. Is this possible?

EDIT: Motivation for analysis of the weighted zeta function

One can relatively easily show that

begin{equation}frac{1}{zeta_a(s)}=sum_{q=1}^{infty}frac{mu(q)mathbb{E}_{ninmathbb{N}}[a_{qn}]}{q^s}tag{3}end{equation}

If $frac{1}{zeta_a(s)}$ has an analytic continuation to $s=1$ and if

begin{equation}lim_{Ntoinfty}frac{1}{N}sum_{q=1}^N mu(q)mathbb{E}_{ninmathbb{N}}[a_{qn}]tag{4}end{equation}

then that is sufficient to show that (3) converges at $s=1$, ensuring the identity

$$sum_{q=1}^{infty}frac{mu(q)mathbb{E}_{ninmathbb{N}}[a_{qn}]}{q}=0$$

for any bounded sequence $a_n$ for any bounded sequence $a_n$. I have ideas on how to show that (4) holds, and so if one could resolve the question of analytic continuation to $s=1$ I would have this nice identity. For specific sequences $a_n$ one gets various interesting results, like for $a_n$ periodic mod $r$ where we get that

$$sum_{q=1}^{infty}frac{mu(q)f(gcd(r,q))}{q}=0$$

for $a_n$ multiplicative we can show that if $f(n)$ is a multiplicative funnction with $mathbb{E}_{ninmathbb{N}}[f(n)]neq0$ then

$$sum_{n=1}^{infty}frac{mu(n)f(n)}{n}=0$$

having a nice identity to tie all these results together would be the dream and it would be nice if we could establish a large chunk of the criterion by resolving this problem.