= Solution
The zero-prescription claim requires a <closed discrete subset>, meaning no accumulation in $\Omega$. If “discrete” means only that each listed point is isolated, it is false: $\{1/n:n\geq1\}\subset\mathbb C$ is a <discrete subset>, but any <holomorphic function> vanishing there is identically zero by the <identity theorem for holomorphic functions>, and then has additional zeros. In the usual zero-set interpretation, discreteness includes local finiteness. We prove the claim with this necessary interpretation.
First choose an increasing <compact exhaustion> $K_j$ of $\Omega$ by relatively filled sets: every component of $\widehat{\mathbb C}\setminus K_j$ meets $\widehat{\mathbb C}\setminus\Omega$. Such an exhaustion is obtained by exhausting with finite unions of small closed squares lying in $\Omega$, then adding all complementary components relatively compact in $\Omega$. The filled sets remain compact in $\Omega$: a complementary component containing a point within distance $\operatorname{dist}(K_j,\mathbb C\setminus\Omega)$ of the complement could join that complement through a ball disjoint from $K_j$, and hence would not be filled. Their boundedness follows by using a disk containing the initial compact. Nesting can be ensured by enlarging each stage first. This is also the geometric description of the <holomorphic convex hull> proved below.
A finite prescribed set is realized by a finite product of linear factors. For an infinite <closed discrete subset>, enumerate the distinct points as $a_n$. Every compact contains only finitely many of them, so there are indices $j(n)\to\infty$ with $a_n\notin K_{j(n)}$. Finitely many initial factors can be handled without an approximation requirement. For the remaining factors, let $U_n$ be the complementary component containing $a_n$. It also contains a point $c_n\notin\Omega$ or infinity. Join $a_n$ to $c_n$ by an arc in that component, or to infinity by a ray-like arc. Define
$$
p_n(z)=\frac{z-a_n}{z-c_n}\quad(c_n\ne\infty),
\qquad p_n(z)=z-a_n\quad(c_n=\infty).
$$
Each $p_n$ is a <holomorphic function> on $\Omega$, has exactly one <simple zero>, at $a_n$, and no other zeros there.
The arc is disjoint from $K_{j(n)}$, and $p_n$ has a <holomorphic logarithm> $L_n$ near that compact. To justify the logarithm, any closed curve avoiding the arc has equal <winding numbers> about its finite endpoints, so the integral of $p_n'/p_n$ around it is zero. For a ray to infinity the winding number about $a_n$ is zero. A primitive of $p_n'/p_n$, adjusted by a constant on each component, supplies the logarithm.
The <Runge theorem> proved in Question 4 approximates $L_n$ uniformly on $K_{j(n)}$ by $h_n\in\mathcal O(\Omega)$: all needed finite poles can be chosen outside $\Omega$ because every complementary component meets its complement. Choose the error small enough that
$$
F_n=p_n e^{-h_n},\qquad \sup_{K_{j(n)}}|F_n-1|<2^{-n}.
$$
Each $F_n$ still has exactly the <simple zero> $a_n$. On every fixed compact, all sufficiently late factors satisfy the summable bound above. Hence their <infinite product> converges locally uniformly. It has a nonzero tail because, once $|F_n-1|<1/2$, its logarithmic tails satisfy $|\log F_n|\leq2|F_n-1|$ and converge absolutely. Thus multiplying by the finitely many earlier factors gives \b[a holomorphic function with precisely the prescribed zeros, and no additional zeros]. Repeating a factor to any prescribed finite positive multiplicity gives the corresponding multiplicity version. This is <prescribed zeros in a plane domain via Runge approximation>, not an invocation of the existence theorem being asked for.
For the specified <infinite product>, put $a_k=q^{-2k-1}$. On any compact annulus $0<r\leq|z|\leq R$,
$$
\sum_{k\geq0}\bigl(|a_kz|+|a_kz^{-1}|\bigr)
\leq(R+r^{-1})\frac{|q|^{-1}}{1-|q|^{-2}}<\infty.
$$
The logarithmic-tail estimate therefore proves locally uniform convergence on $\mathbb C^*$ and nonvanishing wherever no individual factor vanishes. At a zero of one factor, the product is zero and the remaining product converges to a nonzero value. Consequently
$$
\boxed{\text{The product is defined and converges for every }z\ne0;
\quad Z(\sigma)=\{q^{2k+1},q^{-2k-1}:k\geq0\}.}
$$
All these <zeros> are simple and distinct: their moduli distinguish the two strings and their indices. At $z=0$ the reciprocal factors are undefined, so the given expression has no convergence value there. The infinite sequence of zeros tending to zero also rules out continuation across that point as a <holomorphic function>.
Set $P(z)=\prod_{k\geq1}(1-q^{-2k}z)(1-q^{-2k}z^{-1})$. Absolute local convergence allows reindexing:
$$
\sigma(qz)=(1-z)P(z),\qquad
\sigma(q^{-1}z)=(1-z^{-1})P(z).
$$
Since $1-z=-z(1-z^{-1})$, \b[$\sigma(qz)=-z\sigma(q^{-1}z)$ for every $z\ne0$]. No division by a vanishing factor is involved, so the identity also holds at all the zeros.
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