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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: $$Q''-(1/2){Q'}^2=P,$$ so this is a condition for a Riccati equation to have a polynomial solution $Q'$.

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 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, is classical and well-known. It follows from asymptotic expansions of these solutions for large values of the independent variable, or in a simpler way, from the Wiman-Valiron theory.

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: $$Q''-(1/2){Q'}^2=P,$$ so this is a condition for a Riccati equation to have a polynomial solution $Q'$.

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.

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: $$Q''-(1/2){Q'}^2=P,$$ so this is a condition for a Riccati equation to have a polynomial solution $Q'$.

The answer is this: $$f(z)=\int_{z_0}^z e^{P(\zeta)}d\zeta,$$ where $P$ 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 $e^P$.

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.

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.