The other possibility is that $f(z)=e^z$. Explicit computation gives $$\frac{d}{dr}f^\#(re^{i\phi})=2e^{r\cos\phi}\cos\phi(1-e^{2r\cos\phi}),$$ which is $\leq 0$ everywhere.
On the other hand, the exponential is the only exception among entire functions.
Indeed, if $f^\#$ is decreasing on each radius, then $f^\#$ has no zeros, this implies that $f'$ has no zeros. On the other hand, $f^\#(z)\leq f^\#(0)$ must be bounded, and a theorem of Clunie and Hayman implies that $f$ has at most exponential type (that is at most order 1, normal type). Then $f'$ has at most exponential type, and since it has no zeros, it must be $e^{az+b}$ by the Hadamard factorization theorem.
Refs. The original paper of Clunie and Hayman is
MR0192055 Clunie, J.; Hayman, W. K. The spherical derivative of integral and meromorphic functions. Comment. Math. Helv. 40 (1966), 117–148.
Their proof was much simplified (and generalized) in
MR2869124 Barrett, Matthew; Eremenko, Alexandre Generalization of a theorem of Clunie and Hayman. Proc. Amer. Math. Soc. 140 (2012), no. 4, 1397–1402.
Edit. It is probably not very difficult to describe all meromorphic functions with your property. Again your condition implies that $f$ has no critical points, but Clunie-Hayman theorem is not applicable, and we only have that $f$ is of order at most 2, normal type. Still we obtain a very narrow class of functions, and I have no doubt that it can be determined which of them have your property, if this is needed.