Ring of rational-coefficient power series defining entire functions

I'm wondering if anyone has come across the following ring before. Let $R$ be the ring of complex power series $f=\sum_{n \ge 0} a_n t^n$ such that

1. $a_n \in \mathbb{Q} \: \: \forall \: n$
2. The function $f: \mathbb{C} \rightarrow \mathbb{C}$ is entire

Any information about it would be welcome but in order to keep the question specific, I'll list a few things I would like to know in particular.

1. Is $R$ a unique factorisation domain?
2. Are there any elements of $R$ which are known to be prime/irreducible but are not associates of polynomials?
3. If an element $f \in R$ is prime/irreducible, must all its zeroes (in $\mathbb{C}$) be simple?
4. Is every $\alpha \in \mathbb{C}$ a root of some nonzero $f \in R$ ?

Relevant here is the paper of Hurwitz:

Über beständig convergirende Potenzreihen mit rationalen Zahlencoefficienten und vorgeschriebenen Nullstellen. Acta Math. 14 (1890), no. 1, 211–215.

Since the results of that paper are simple and short, I will explain them here. Let $f$ be any entire function. Hurwitz proves that there exists another entire function $g$ such that $f \cdot e^{g}$ (which is entire) has coefficients in $\mathbf{Q}(i)$. The proof is quite easy. Assume (for convenience, one can reduce to this case) case that $f(0) \ne 0$. Then $\log(f(z))$ is holomorphic in a neighbourhood of zero, and so one can write:

$$\log f(z) = b_0 + b_1 z + b_2 z^2 + \ldots $$

Since $\mathbf{Q}(i)$ is dense in $\mathbf{C}$, there exists a function $h(z) = a_0 + a_1 z + a_2 z^2 + \ldots $ so that:

1. $a_0 = 0$,
2. $g(z) = h(z) - \log f(z)$ is entire,
3. $h(z)$ has coefficients in $\mathbf{Q}(i)$.

For example, one can take $a_k$ for $k \ge 1$ to be $1/k!$ times the nearest Gaussian integer to $k! \cdot b_k \in \mathbf{C}$, so $|a_k - b_k| < 1/k!$ is decreasing fast enough to ensure that $g(z)$ is entire. It follows that

$$f \cdot e^g = e^{h(z)}.$$

By construction, the function $h(z)$ has trivial constant term and coefficients in $\mathbf{Q}(i)$. It follows easily that so does $f \cdot e^g$. Now we also observe that $f \cdot e^g$ has the same zeros as $f$. So the properties of ideals in the ring of entire functions with coefficients in $\mathbf{Q}(i)$ is very similar to the properties of ideals in the ring of entire functions. Similarly, the ring of entire functions with coefficients in $\mathbf{Q}$ is very similar to the properties of ideals in the ring of entire functions with coefficients in $\mathbf{R}$ (equivalently, entire functions with $\overline{f(z)} =f(\overline{z})$).

Now recall:

1. If $S \subset \mathbf{C}$ is any subset of $\mathbf{C}$ which is closed and discrete, there exists an entire function with zeros exactly at $S$ to any prescribed multiplicity. (One can simply write down such a function as an infinite product.)

2. If $S \subset \mathbf{C}$ is any subset as above such that $\overline{S} = S$, then there exists an entire function with zeros exactly at $S$ to any prescribed multiplicity with coefficients in $\mathbf{R}$, as long as the multiplicity at $s$ is the same as at $\overline{s}$. By Part I, there exists a complex entire function $f = \sum a_n z^n$. The conjugate $\overline{f}(z) = \sum \overline{a_n} z^n$ is a second function. It follows that $$\sqrt{f(z) \cdot \overline{f}(z)}$$ is the required function. (This squareroot exists as an entire function because the order of every zero of $f(z) \cdot \overline{f}(z)$ is even.)

In particular, Hurwitz proves that:
1. If $\alpha$ is any real number, there exists an entire function with rational coefficients that has a simple zero at $\alpha$ and nowhere else.
2. If $\beta$ is any complex number, there exists an entire function with rational coefficients that has a simple zero at $\beta$ and $\overline{\beta}$ and nowhere else.

The construction of these functions can even be said to be quite explicit.

Functions of type 1. and 2. are both prime/irreducible, and consist of all the irreducible elements up to units. (Clearly if $f(z)$ is not a unit, it either has a real zero or a pair of conjugate complex zeros, and so is divisible by such a function.)

We deduce:

1. $R$ is not a UFD, because $\sin(z)$ does not factor into irreducibles (any factorization of $\sin(z)$ into finitely many factors has to contain at least one function with infinitely many zeros, which is not irreducible.)

2. Yes; any function of type 1 or 2 where $\alpha$ is not rational or $\beta$ is not a quadratic imaginary will do.

3. Yes, by the above, if $f(z)$ is irreducible, not only are the zeros of $f(z)$ simple, but there are at most two of them! (In particular, most "irreducible polynomials" are not irreducible in $R$.)

4. Yes, there is even such a function with zeros only at $\alpha$ and $\overline{\alpha}$.

Some extra facts which are easy to deduce from this construction: $R$ is a Bezout domain: any finitely generated ideal is principal.

A start, but perhaps better than nothing:

1. If $r$ is a rational number, then $\sin(r t)$ and $\cos(r t)$ are elements of your ring. There is an identity

$$\sin(r t) = 2 \sin(r/2 t) \cos(r/2 t).$$

Neither element on the right is a unit. This implies that the sequence of ideals:

$$ (\sin(t)) \subsetneq (\sin(t/2)) \subsetneq (\sin(t/4)) \subsetneq \cdots $$

is an infinite ascending chain of distinct ideals. So $R$ is not Noetherian (and hence definitely not a PID).