= Solution
The complement of $D$ is nonempty and closed. Take points $w_j$ in that complement with $|z_0-w_j|\to\delta(z_0)$. This sequence is bounded, so a subsequence converges to a point $w_0\notin D$. Continuity of distance gives $|z_0-w_0|=\delta(z_0)>0$.
Put $t=(z_0-w_0)/(z-w_0)$. On the indicated exterior region $|t|<1$, and $f=1-t$ lies in the right half-plane. Its principal <holomorphic logarithm> is therefore the <power series>
$$
\log f(z)=-\sum_{k=1}^{\infty}\frac{t^k}{k}.
$$
On each compact subset of that region, $|t|\leq q<1$, so the series has <locally uniform convergence> by comparison with $\sum q^k/k$. It does \b[not converge uniformly on the whole region]: choose $z=w_0+(z_0-w_0)/s$ with $0<s<1$. Then $t=s$; as $s\uparrow1$, $\log(1-s)$ tends to minus infinity while each fixed partial sum stays bounded. In particular, the remainder after any fixed number of terms is unbounded.
The <boundary-adapted holomorphic zero factor> $E_0$ is holomorphic throughout $D$, since its only possible denominator singularity is at $w_0\notin D$. The exponential never vanishes, so its only zero is $z_0$, and
$$
E_0'(z_0)=\frac{1}{z_0-w_0}\exp\left(\sum_{k=1}^K\frac1k\right)\ne0.
$$
On $|z-w_0|>2\delta(z_0)$, cancellation of the first $K$ logarithmic terms gives a <holomorphic logarithm> of this zero-free restriction with
$$
\boxed{\left|\log E_0(z)\right|\leq\sum_{k>K}\frac{2^{-k}}k\leq\frac{2^{-K}}{K+1}.}
$$
Choose $K$ so that the final bound is smaller than the prescribed error. This logarithm is only asserted on the exterior region; a function with a zero cannot have a logarithm on all of $D$.
For each $z_n$, construct a <boundary-adapted holomorphic zero factor> $E_n$ whose logarithmic error is smaller than $2^{-n}$. If $L\subset D$ is compact, write $d_L=\operatorname{dist}(L,\mathbb C\setminus D)>0$. For all sufficiently large $n$, $2\delta(z_n)<d_L\leq|z-w_n|$ for every $z\in L$. Hence $\sum\log E_n$ converges absolutely and uniformly on $L$ after discarding finitely many terms. Consequently
$$
\boxed{F(z)=\prod_{n=1}^{\infty}E_n(z)}
$$
is a <holomorphic function>. Its tail is the exponential of a convergent logarithmic sum and is nonzero. The finite initial product has precisely its prescribed <simple zeros>. Since $\delta(z_n)\to0$, only finitely many $z_n$ lie in any compact subset of $D$, and this proves that $F$ has exactly the desired <simple zeros>.
For the final <meromorphic function> assertion, poles escaping to infinity need additional care: a locally finite pole sequence need not satisfy $\delta(z_n)\to0$. Here is the required extension of the product argument. Split the distinct poles $a$ into two sets according as
$$
\delta(a)\leq\frac{1}{1+|a|}\quad\hbox{or}\quad\delta(a)>\frac{1}{1+|a|}.
$$
In the first set, for each $\varepsilon>0$ the poles with $\delta(a)\geq\varepsilon$ are in a bounded closed subset lying at least $\varepsilon$ from the complement. Local finiteness makes this a finite set. Thus the boundary distances tend to zero. In the second set, poles in $|a|\leq R$ are at least $1/(1+R)$ from the complement, so there are finitely many of them. Thus their moduli tend to infinity.
Use the preceding factors for the first set. For the second set use the <Weierstrass elementary factors>
$$
P_K(z/a)=(1-z/a)\exp\left(\sum_{k=1}^K\frac{(z/a)^k}{k}\right).
$$
For $|z|<|a|/2$ these have the same logarithmic-tail bound as above. If a pole has order $m$, choose the factor's error smaller than $2^{-n}/m$ and raise it to the power $m$. A possible pole at zero is handled by a finite polynomial factor. On every compact subset the resulting logarithmic tails are uniformly summable, so their product $q$ is holomorphic with zeros of exactly the pole orders and no other zeros. If $M$ is the original <meromorphic function>, the product $Mq$ extends holomorphically across every pole. Writing that extension as $p$ gives the concise conclusion
$$
\boxed{M=p/q,\qquad p,q\text{ holomorphic on }D,\quad q\not\equiv0.}
$$
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