If an infinite product \[ \prod_{n} (1 + a_{n}(s)) \] converges, then can we say anything about its lower bound?
Or in general, can it take any value arbitrarily close to $0$?
What if $a_{n}$ is of the form $a_{n}(s) = b_{n}m_{n}^{-s}$, where $\{m_{n} \}$ is a subsequence of $\{1, 2, \ldots \}$ and |b_{n}| = 1?
The product can take an arbitrary value, in particular any value arbitrarily close to zero. Take any infinite product that converges to some value $\alpha$. To let it converge to arbitrary $\beta\in\mathbb C$ instead, multiply the first factor by $\beta/\alpha$. In terms of your factors $1+a_n(s)$, you need
$$ 1+a_n'(s)=\frac\beta\alpha(1+a_n(s)) $$
and thus
$$ a_n'(s)=\frac\beta\alpha(1+a_n(s))-1\;. $$
Edit in response to the new question:
The product can still be zero if $b_1=-1$ and $m_1=1$. Excluding this trivial case, whether the product can converge to zero now depends on $s$.
Since the product converges, $a_n$ converges to $0$; it follows that $s\gt0$. Then the product is minimal if $b_n=-1$ and $m_n=n$ for all $n\ge2$, yielding
$$ \prod_{n\ge2}\left(1-n^{-s}\right)\;, $$
with another factor of $2$ if we choose to include $m_1=1$ with $b_1=1$. We can obtain a lower bound as follows:
\begin{eqnarray*} \prod_{n\ge2}\left(1-n^{-s}\right) &=& \exp\left(\sum_{n\ge2}\log\left(1-n^{-s}\right)\right)\\ &\lt& \exp\left(-\sum_{n\ge2}n^{-s}\right)\\ &=&\exp\left(1-\zeta(s)\right)\;. \end{eqnarray*}
This holds for $s\gt1$. The zeta function has a pole at $s=1$, and in this case the sum in the exponent diverges logarithmically as $n\to\infty$. Thus, for $s\le1$ the product can converge to zero.