Your question falls under the more general class of mappings of the form $$f_{\lambda}(z)=z+\lambda \sin(z), $$ for a parameter $\lambda$. Discussion about such functions may be found in the article Wandering domains in the iteration of entire functions-Baker, where a general approach to similar questions is described on page 567, using a logarithmic change of variables and an example quite reminiscent of yours is worked out on page 569. More information may be found in the following article of Mr. Bergweiler, "Iteration of meromorphic functions", for which I am unable to provide a link, on page 168.
In my opinion though, if what you are looking for is just a concrete example of an entire function possessing a wandering domain, the mapping $$f(z)=z+2\pi + \sin(z)$$ is much simpler to work with. That being said, it requires some estimates on the growth of entire functions on points belonging in the Fatou set. Once again, for the aforementioned estimates I will cite Mr. Bergweiler's article, page 165, Lemma 7.
This particular example is worked out in The Teichmüller space of an entire function-Fagella, Henriksen, on page 21. Lets see how wandering domains arise here.
Observe that the critical points of $f$ are precisely at the points $c_{k}=(2k+1)\pi$, for any integer $k$. In a sense, these points wander, and as it will become apparent, those $c_{k}$ will be the "centers" of the wandering domain. Note that $$f(c_{k})=c_{k+1},$$ while the lines $l_{k}$ situated at $\operatorname{Re(z)=2k \pi}$ are subsets of the corresponding Julia set and $$f(l_{k})=l_{k+1}.$$ Here the estimates on growth I mentioned above come to the rescue in order to justify this claim. These will serve as a barier so that, the wandering domain is contained in the vertical strip defined by two such lines. The dynamical behaviour of $f$ may be witnessed in the following Image. This is an image from the article of Ms. Fagella and Mr. Henriksen, mentioned above.
We may perform a semi-conjugation with the exponential $E(z)=e^{i z}$. Then, $E\circ f=F\circ E$, where $$F(z)=z \exp \left\{ \frac{1}{2} \left(z-\frac{1}{z} \right) \right\},$$ much like your case. Note that $F$ has a superattracting fixed point at $z=-1$, though care should be exercised, as this is a mapping of Radstrom type, On the iteration of analytic functions-Radstrom. Furthermore, the point $z=-1$ is the only critical point contained in its immediate basin of attraction, so by the standrad theory it is a simply connected domain and all critical points $c_{k}$ for $f$ lie over $z=-1$ under the semi-conjugation. As a consequence, the logarithmic image of the basin of $z=-1$ produces an infinitude of simply connected components of the Fatou set for $f$, each one containing a point $c_{k}$. As the points $c_{k}$ wander, so do these components.
In case you wonder, if there are many entire functions with wandering domains, it has been proved that to every transcendental entire function $f$ having a fixed point, corresponds another entire function $g$, so that $f\circ g$ possesses a wandering domain. The assumption on the fixed point may be removed as, for any transcendental entire function $f$, the composition $f\circ f$ has a fixed point, (Picard's little theorem). This result is due to Wandering domains in the iteration of compositions of entire functions-Baker, Singh.
I hope this will be helpful.
Best Answer
If $f$ is a non-linear polynomial, then $A(\infty)$ is open (to prove it, try to prove that there exists $R>0$ such that $A(\infty)=\bigcup_n f^{-n}\{ z \in \mathbb C : |z|>R \}$.
If $f$ is entire and not a polynomial, then as Martin R pointed out the escaping set $I(f)$ is not open (usually the term "basin" is reserved for open sets) and is not contained in the Fatou set.