Mellin transform of function $f(x)$ which is continuous and $f(x)\geq 0 $ is given as $$ f^\ast(z) =\int\limits_0^\infty x^{z} f(x) \frac{dx}{x}. $$ 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 . 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 $f^\ast(z)=\Gamma(z)$.
The Mellin transform of $1/(e^x+1)$ is $\Gamma(s)\eta(s), \Re s >0$, where as usual $\eta(s)=(1-2^{1-s})\zeta(s)$ which has tons of zeroes both on the line $\Re s =1$ and on the line $ \Re s =1/2$ (and of course possibly more depending on RH) so the OP question has a negative answer.
Not sure if there is any simple relation between the function $f$ and the zeroes (or more generally value distribution) of $f^*$