$\newcommand\R{\mathbb R}\newcommand\Z{\mathbb Z}$After [James Martin's clarifying comment][1], the question becomes as follows: 

>Suppose that $Z=XY$, where $X$ and $Y$ are independent positive random variables (r.v.'s). The distributions $P_X$ and $P_Z$ of $X$ and $Z$ are known. Does this determine the distribution $P_Y$ of $Y$? 

The answer to this question is no. Indeed, write $X=e^U$, $Y=e^V$, and $Z=e^W$, where $U,V,W$ are real-valued r.v.'s such that $W=U+V$; $U$ and $V$ are independent; and the distributions $P_U$ and $P_W$ are known. The question can now be restated in terms of the characteristic functions (c.f.'s) $f_U,f_V,f_W$ of $U,V,W$ as follows:

>Suppose that $f_U\,f_V=f_W$ and suppose that $f_U$ and $f_W$ are known. Does this determine $f_V$?

The answer to this equivalent question is of course still no. Indeed, note that (i) the function $f$ given by the formula $f(t)=\max(0,1-|t|/\pi)$ for real $t$ is a c.f. (of the absolutely continuous distribution with density $\R\ni x\mapsto\dfrac{1-\cos\pi x}{\pi^2 x^2}\,1(x\ne0)$ and (ii) the periodic function $g$ with period $2\pi$ such that $g=f$ on $[-\pi,\pi]$ is a c.f. (of the discrete distribution on $\Z$ with probability mass function 
$\Z\ni x\mapsto\dfrac12\,1(x=0)+\dfrac{1-\cos\pi x}{\pi^2 x^2}\,1(x\ne0)$. Note that $fg=f^2$. So, if $f_U=f$ and $f_W=f^2$, then $f_V$ can be either one of the two distinct c.f.'s: $f$ or $g$. So, $f_U$ and $f_W$ do not determine $f_V$. $\quad\Box$.

  [1]: https://mathoverflow.net/questions/446738/approximation-to-ratio-distribution?noredirect=1#comment1154351_446738