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 $P(z)$.
Best Answer
Let $f$ be a meromorphic function on $\mathbb{C}$. Assume that $f$ does omit the different values $a,b,c \in \mathbb{C}$. Then $f-a$ does not assume $0$, and $1/(f-a)$ extends to an entire function (the poles of $f$ are the zeros of $1/(f-a)$). Now $1/(f-a)$ omits the values $1/(b-a)$ and $1/(c-a)$. According to Little Picard's Theorem $1/(f-a)$ and therefore $f$ is constant. Thus a nonconstant meromorphic function on $\mathbb{C}$ can omit at most 2 values. The function $z \mapsto \tan(z)$ for example is meromorphic on $\mathbb{C}$ and omits $\pm i$, since $$ 1+\tan^2(z)= \frac{1}{\cos^2(z)} $$