Mellin transform of function $f(x)$ defined for $x\geqslant 0$ is given by $$ f^\ast(z) =\int\limits_0^\infty x^{z} f(x) \frac{dx}{x}. $$ I consider only exponentially decreasing (there exist such constants $C_1>0$, $C_2>0$ that $|f(x)| \leqslant C_1 e^{-C_2x}$), strictly positive, continuous on $[0,+\infty)$ and infinitely many times differentiable on $(0,+\infty)$ functions $f(x)$ such that all its derivatives decrease exponentially at infinity (and even with the same $C_2$). For such functions $f^\ast(z)$ is defined and analytic for $\Re(z) > 0$.
Question 1. Is it possible to say something about existence of zeros of $f^\ast(z)$ in $\Re(z)$>0? Is it true that $f^\ast(z)$ can't have zeros in $\Re(z)>0$? If it can have zeros is it known which conditions on $f(x)$ will garantee the absense of zeros of $f^\ast(z)$? Motivating example is $f(x)=e^{-x}$ with $f^\ast(z)=\Gamma(z)$.
Question 2. Is it possible to say something about decreasing rate (lower and upper bounds) of $f^\ast(z)$ along vertical lines $\Re(z) = \mathrm{const}$? Motivating example is again $f(x)=e^{-x}$ with $f^\ast(z)=\Gamma(z)$. Here $f^\ast(z)$ decreases exponentially with order $\frac{\pi}{2}$ for $\Re(z)\in(0,\frac{1}{2}]$ and faster then any exponent of order less that $\frac{\pi}{2}$ for $x>\frac{1}{2}$.