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(ii) If we only have an i.i.d. sample $X_1,\dots,X_n$ from the individual distribution of $X$ and an i.i.d. sample $Z_1,\dots,Z_n$ from the individual distribution of $Z$, then we only have partial knowledge of the individual distributions of $X$ and and of $Z$. But, as shown above, even complete knowledge of the individual distributions of $X$ and and of $Z$ will in general not determine the distribution of $Y$. So, clearly, the partial knowledge of the individual distributions of $X$ and and of $Z$ cannot provide substantial partial knowledge of the distribution of $Y$.

(ii) If we only have an i.i.d. sample $X_1,\dots,X_n$ from the individual distribution of $X$ and an i.i.d. sample $Z_1,\dots,Z_n$ from the individual distribution of $Z$, then we only have partial knowledge of the individual distributions of $X$ and and of $Z$. But, as shown above, even complete knowledge of the individual distributions of $X$ and of $Z$ will in general not determine the distribution of $Y$. So, clearly, the partial knowledge of the individual distributions of $X$ and and of $Z$ cannot provide substantial partial knowledge of the distribution of $Y$ -- in particular, then we would even be unable to guess whether $Y$ is integer-valued or absolutely continuous.

Summarizing the statistical sampling aspects of the problem: depending on the kind of of sampling available -- (i) from the joint distribution of $(X,Z)$ or (ii) from the individual distributions of $X$ and $Z$, the sampling is either (i) irrelevant to the problem stated in the OP or (ii) not at all helpful to the problem.

$\newcommand\R{\mathbb R}\newcommand\Z{\mathbb Z}$After James Martin's clarifying comment, 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 $f_V$. $\quad\Box$.

This example is well known; see e.g. this book.

Such examples are of course an exception. Indeed, if the set $N_U:=\{t\in\R\colon f_U(t)=0\}$ is nowhere dense (as will usually be the case, apparently -- including the case when $f_U$ is analytic in a neighborhood of $\R$), then we know the values of $f_V(t)=f_W(t)/f_U(t)$ for all $t$ in the everywhere dense set $\R\setminus N_U$. So, by the continuity of any c.f., we know $f_V$ completely -- so that we know the distribution of $V$ and hence the distribution of $Y$.

Concerning statistical aspects of the problem, raised by James Martin, one can say the following.

(i) If we have an i.i.d. sample $(X_1,Z_1),\dots,(X_n,Z_n)$ from the joint distribution of the pair $(X,Y)$ for a large enough sample size $n$, we get the i.i.d. $Y$-sample $Y_1:=Z_1/X_1,\dots,Y_n:=Z_n/X_n$. So, we can use, say, the empirical c.d.f. based on $Y_1,\dots,Y_n$ to approximate in the standard manner the true c.d.f. of $Y$. The question as to whether the distribution of $Y$ is determined by the individual distributions of $X$ and and of $Z$ is completely irrelevant here, once the joint distribution of the pair $(X,Y)$ is known or available for sampling as described above.

(ii) If we only have an i.i.d. sample $X_1,\dots,X_n$ from the individual distribution of $X$ and an i.i.d. sample $Z_1,\dots,Z_n$ from the individual distribution of $Z$, then we only have partial knowledge of the individual distributions of $X$ and and of $Z$. But, as shown above, even complete knowledge of the individual distributions of $X$ and and of $Z$ will in general not determine the distribution of $Y$. So, clearly, the partial knowledge of the individual distributions of $X$ and and of $Z$ cannot provide substantial partial knowledge of the distribution of $Y$.

$\newcommand\R{\mathbb R}\newcommand\Z{\mathbb Z}$After James Martin's clarifying comment, 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 $f_V$. $\quad\Box$.

This example is well known; see e.g. this book.

Such examples are of course an exception. Indeed, if the set $N_U:=\{t\in\R\colon f_U(t)=0\}$ is nowhere dense (as will usually be the case, apparently -- including the case when $f_U$ is analytic in a neighborhood of $\R$), then we know the values of $f_V(t)=f_W(t)/f_U(t)$ for all $t$ in the everywhere dense set $\R\setminus N_U$. So, by the continuity of any c.f., we know $f_V$ completely -- so that we know the distribution of $V$ and hence the distribution of $Y$.

Regarding concerns about statistical aspects of the problem, raised by James Martin, one can say the following.

(i) If we have an i.i.d. sample $(X_1,Z_1),\dots,(X_n,Z_n)$ from the joint distribution of the pair $(X,Y)$ for a large enough sample size $n$, we get the i.i.d. $Y$-sample $Y_1:=Z_1/X_1,\dots,Y_n:=Z_n/X_n$. So, we can use, say, the empirical c.d.f. based on $Y_1,\dots,Y_n$ to approximate in the standard manner the true c.d.f. of $Y$. The question as to whether the distribution of $Y$ is determined by the individual distributions of $X$ and and of $Z$ is completely irrelevant here, once the joint distribution of the pair $(X,Y)$ is known or available for sampling as described above.

(ii) If we only have an i.i.d. sample $X_1,\dots,X_n$ from the individual distribution of $X$ and an i.i.d. sample $Z_1,\dots,Z_n$ from the individual distribution of $Z$, then we only have partial knowledge of the individual distributions of $X$ and and of $Z$. But, as shown above, even complete knowledge of the individual distributions of $X$ and and of $Z$ will in general not determine the distribution of $Y$. So, clearly, the partial knowledge of the individual distributions of $X$ and and of $Z$ cannot provide substantial partial knowledge of the distribution of $Y$.

Such examples are of course an exception. Indeed, if the set $N_U:=\{t\in\R\colon f_U(t)=0\}$ is nowhere dense (as will usually be the case, apparently -- including the case when $f_U$ is analytic in a neighborhood of $\R$), then we know the values of $f_V(t)=f_W(t)/f_U(t)$ for all $t$ in the everywhere dense set $\R\setminus N_U$. So, by the continuity of any c.f., we know $f_V$ completely -- so that we know the distribution of $V$ and hence the distribution of $Y$. Given now values of $(X_1,Z_1),\dots,(X_n,Z_n)$ of i.i.d. copies of the pair $(X,Y)$ for a large enough sample size $n$, we get the i.i.d. $Y$-sample $Y_1:=Z_1/X_1,\dots,Y_n:=Z_n/X_n$. So, we can use, say, the empirical c.d.f. based on $Y_1,\dots,Y_n$ to approximate in the standard manner the true c.d.f. of $Y$.

Such examples are of course an exception. Indeed, if the set $N_U:=\{t\in\R\colon f_U(t)=0\}$ is nowhere dense (as will usually be the case, apparently -- including the case when $f_U$ is analytic in a neighborhood of $\R$), then we know the values of $f_V(t)=f_W(t)/f_U(t)$ for all $t$ in the everywhere dense set $\R\setminus N_U$. So, by the continuity of any c.f., we know $f_V$ completely -- so that we know the distribution of $V$ and hence the distribution of $Y$.

Concerning statistical aspects of the problem, raised by James Martin, one can say the following.

(i) If we have an i.i.d. sample $(X_1,Z_1),\dots,(X_n,Z_n)$ from the joint distribution of the pair $(X,Y)$ for a large enough sample size $n$, we get the i.i.d. $Y$-sample $Y_1:=Z_1/X_1,\dots,Y_n:=Z_n/X_n$. So, we can use, say, the empirical c.d.f. based on $Y_1,\dots,Y_n$ to approximate in the standard manner the true c.d.f. of $Y$. The question as to whether the distribution of $Y$ is determined by the individual distributions of $X$ and and of $Z$ is completely irrelevant here, once the joint distribution of the pair $(X,Y)$ is known or available for sampling as described above.

(ii) If we only have an i.i.d. sample $X_1,\dots,X_n$ from the individual distribution of $X$ and an i.i.d. sample $Z_1,\dots,Z_n$ from the individual distribution of $Z$, then we only have partial knowledge of the individual distributions of $X$ and and of $Z$. But, as shown above, even complete knowledge of the individual distributions of $X$ and and of $Z$ will in general not determine the distribution of $Y$. So, clearly, the partial knowledge of the individual distributions of $X$ and and of $Z$ cannot provide substantial partial knowledge of the distribution of $Y$.