= Solution
For <infinite products>, a nonzero limiting product requires that its factors tend to one. A useful sufficient condition is
$$
\sum_j|u_j|<\infty,\qquad 1+u_j\ne0.
$$
After finitely many factors, $|u_j|<1/2$ and the principal <holomorphic logarithm> satisfies $|\log(1+u_j)|\leq2|u_j|$. Therefore the sum of logarithms converges, and exponentiating it gives a finite nonzero product. The same argument on compact sets proves <infinite product convergence from logarithmic tails>: a locally uniformly absolutely convergent tail of <holomorphic> logarithms gives a <holomorphic> nonvanishing tail product. Finite factors then determine all zeros and their orders.
The <Weierstrass elementary factors> are
$$
E_p(w)=(1-w)\exp\left(w+\frac{w^2}{2}+\cdots+\frac{w^p}{p}\right),\quad
E_0(w)=1-w.
$$
For $|w|<1$,
$$
\log E_p(w)=-\sum_{\nu=p+1}^{\infty}\frac{w^\nu}{\nu},
\qquad
|\log E_p(w)|\leq2|w|^{p+1}\quad(|w|\leq1/2).
$$
Their only zero is a simple zero at $w=1$.
Work with <entire functions> on $\mathbb C$. As usual for prescribed exact zero orders, the distinct zero locations must have consistent multiplicities. The literal statement allows repeated locations with conflicting orders; that cannot be true, for example if the same point is prescribed order one and order two. Remove consistent repetitions rather than adding their orders, and separate a possible zero at the origin.
List the distinct nonzero locations as $a_j$, with prescribed orders $n_j$, and write $m$ for the prescribed order at zero, or zero if the origin is not prescribed. Choose $p_j$ large enough that
$$
n_j2^{-p_j}\leq2^{-j}.
$$
Then the <Weierstrass factorization theorem> construction is
$$
\boxed{F(z)=z^m\prod_j E_{p_j}(z/a_j)^{n_j}.}
$$
On every compact set, $|z/a_j|\leq1/2$ for all sufficiently large $j$ because $|a_j|\to\infty$. Its logarithmic tail is bounded by
$$
\sum_j n_j|\log E_{p_j}(z/a_j)|\leq\sum_j n_j2^{-p_j}
\leq\sum_j2^{-j}.
$$
Thus the product converges locally uniformly, is entire, and has exactly the specified zeros with exactly their orders. At a prescribed zero, only its own finite factor vanishes; all other finite factors and the tail product are nonzero.
Without escape to infinity the conclusion fails in general. Distinct proposed zeros $1/j$ accumulate at zero. The <identity theorem> would force an <entire function> with those zeros to vanish identically, which does not have precisely the prescribed isolated zeros. A finite zero set can of course be realized by a polynomial.
For <entire functions with the same zero divisor>, the quotient $q=f/g$ extends through every common zero by cancelling equal local powers. It is entire and nowhere zero. Hence $q'/q$ is entire and has a primitive on the <simply connected> plane. Choose $h(0)$ with $e^{h(0)}=q(0)$ and put
$$
h(z)=h(0)+\int_0^z\frac{q'(\zeta)}{q(\zeta)}\,d\zeta.
$$
The derivative of $qe^{-h}$ vanishes, and its value at zero is one. Therefore
$$
\boxed{f=e^h g.}
$$
If also $f=e^k g$, then $e^{h-k}=1$ everywhere. The continuous difference $h-k$ takes values in the discrete set $2\pi i\mathbb Z$ and so is constant on the <connected> plane:
$$
\boxed{h-k=2\pi i\ell\quad\text{for one fixed }\ell\in\mathbb Z.}
$$
The whole-plane hypothesis matters: a zero-free <holomorphic> quotient on a multiply <connected> domain need not possess a global <holomorphic logarithm>.
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