We know by Picard's theorem that any entire function is either constant, or surjective or misses only 1 point.
It is easy to observe that $\sin{z}$, $\cos{z}$ are surjective.
Is $f \cdot g$ surjective if $f$ and $g$ are entire and surjective? It is indeed true when $f$ and $g$ are polynomials.
Does there a characterization of entire functions missing a point and surjective entire functions? Or any relation between them?
For any non-constant entire function that satisfy: $$\overline{f(z)} = f(\bar{z}) \quad\text{ and }\quad f(\mathbb{R}) = \mathbb{R}\tag{*}$$ $f$ will be a surjection.
If $f$ is a polynomial, then it is surjective by fundamental theorem of algebra. If $f$ is neither a polynomial nor surjective, then by Picard, $f$ avoids a unique $\alpha \in \mathbb{C}$. Since $\overline{f(z)} = f(\bar{z})$, $f$ will also avoid $\bar{\alpha}$. Since $\alpha$ is unique, $\bar{\alpha} = \alpha \implies \alpha \in \mathbb{R}$. This contradicts with the assumption $f(\mathbb{R}) = \mathbb{R}$.
It is easy to see the function $z \sin z$ satisfy $(*)$, so it is a surjection.
Update
Here is an alternate proof using Rouché's theorem. We will prove a slightly stronger statement:
$$(z + \alpha) \sin z \quad\text{ is surjective for any }\;\; \alpha \in \mathbb{C}$$
For $k \in \mathbb{N}$, let $L_k$ be the contour:
$$L_k = \left\{ x + i y \in \mathbb{C} : \max(|x|,|y|) = (k+\frac12)\pi \right\}$$
Along the edges of $L_k$, we have: $$ |\sin(x+iy)| \ge \begin{cases} \cosh(y),& \text{ for } |x| = (k+\frac12)\pi\\ \sinh((k+\frac12)\pi),&\text{ for } |y| = (k+\frac12)\pi\end{cases} \implies |\sin(x+iy)| \ge 1$$
For any $\beta \in \mathbb{C}$, if we choose a $k \in \mathbb{N}$ such that $( k + \frac12 ) \pi > |\alpha| + |\beta|$, then on $L_k$, we have:
$$|z \sin z| - | \alpha \sin z - \beta | \ge ( |z| - |\alpha| ) |\sin z| - |\beta| \ge |z| - |\alpha| - |\beta| > 0$$
By Rouché's theorem, $z\sin z$ and $(z+\alpha)\sin z - \beta$ has same numbers of root within $L_k$.
Since $z = 0$ is a root of $z \sin z$ within
$L_k$, $(z + \alpha) \sin z = \beta$ has a solution within $L_k$.
Since $\beta$ is arbitrary, this implies $(z + \alpha) \sin z$ is surjective.
If one look at the proof carefully, one will realize a similar approach will allow one to show $P(z) \sin z$ is surjective for any non-constant polynomials $P(z)$.
I'll give an example of (surjective)(surjective) = (not surjective) based on a comment by Hagen von Eitzen. $$z\cdot \frac{\exp(z^2)-1}{z}=\exp(z^2)-1$$ Clearly, the function on the right never attains $-1$. It remains to show that $f(z)=\frac{\exp(z^2)-1}{z}$, extended with $f(0)=0$, is surjective.
Any odd entire function is surjective (or identically zero).
Indeed, $f(-z)=-f(z)$ implies that the range of $\mathbb C$ is symmetric about $0$. So, if $f$ omits some nonzero value $w$, it has to omit $-w$ too, contradicting Picard.
To answer the question from the title: For $x, y \in \mathbb{R}$ and $z = x + iy$ $$\left|\sin(z)\right|^2 = \sin^2(x) + \sinh^2(y).$$
Let $k \in \mathbb{N}$ and let $L$ (depending on $k$) be the rectangle
$$L = \{ x + iy \in \mathbb{C} \mid \max \left(|x|, y\sinh(y) \right) = \left(k+\tfrac{1}{2}\right)\pi \}.$$
Then for all $z \in L$ we have $\left|z \sin(z)\right| \geq (k + \frac{1}{2}) \pi$. Now suppose that $z \sin(z)$ avoids the value $w$. Then $$g(z) = \frac{1}{z \sin(z) - w}$$ is entire. Take $k \in \mathbb{N}$ such that $(k + \frac{1}{2}) \pi > |w|$. Then for all $z\in L$ $$ |g(z)| \leq \frac{1}{(k + \frac{1}{2}) \pi - |w|}$$ and by the maximum modulus principle this inequality must also hold inside $L$. Take $k \to \infty$ to conclude that $g = 0$, which is clearly false. Therefore $z \sin(z)$ must be surjective.
