Define property $A$ for an entire function $f(z)$ as
$1)$ $f(z)=0$ has exactly one solution being $z=0$
$2)$ $f(z)=z$ has exactly one solution $=>z=0$ (follows from $1)$ )
$3)$ $f(z)$ is not a polynomial.
Define property $B$ for an entire function $f(z)$ as
$1)$ $f(z)=f_1(z)$ with property $A$.
$2)$ $f_i(z)= ln(f_{i-1}(z)/z)$ for every positive integer $i$.
$3)$ $f_i(z)$ is entire and has property $A$ for every positive integer $i$.
I think there are no entire functions $f(z)$ that have property $B$.
Is that true ?
There is not even an entire function with property $A$.
Suppose $f$ had property $A$. Then $g(z) = \frac{f(z)}{z}$ is an entire function without zeros, not a polynomial, and $g(z) \neq 1$ for $z \neq 0$.
$\infty$ is an essential singularity of $g$ - since $g$ is not a polynomial, hence by the big theorem of Picard, $g$ assumes every complex value with one possible exception infinitely often in every set $\{ z : \lvert z\rvert > K\}$. But property $A$ demands that the two values $0$ and $1$ are omitted.