The answer is this:
$$f(z)=\int_{z_0}^z e^{Q(\zeta)}d\zeta,$$
where $Q$ is a polynomial, and this is the general form of
an entire function whose Schwarzian is a polynomial. The crucial fact that $f$ has finite order.
Then $f'$, a function of finite order without zeros must be of the form $e^Q$.

This is a special case of the theorem of R. Nevanlinna which characterizes
meromorphic solutions of the Schwarz equation
$Sf=P$, where $P$ is a polynomial. The Schwarz differential equation is equivalent to the linear differential equation
$$w''+(P/2)w=0,$$
namely, the general solution $f=w_1/w_2$, where $f_1,f_2$ are two
linearly independent solutions of the linear differential equation. 
And the fact that all solution of the linear differential equation with
polynomial coefficient are entire functions of finite order have finite
order, is classical and well-known.

Ref. R. Nevanlinna, Analytic functions, Chap.XI,3. 

Remark. For generic $P$, the general solution $f$ of $Sf=P$ is
meromorphic but not entire. The condition that it has an entre solution is somewhat complicated.