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Suppose that the set $S:=\{x>0\colon P_X(x)\ne0\}$ has a nonempty intersection with each right neighborhood of $0$.
In terms of the characteristic functions (ch.f.'s) $f_C$ and $f_R$ of $C$ and $R$, your convolution equation means that $g_x(t):=f_C(t)f_R(xt)$ is known for all $x\in S$ and all real $t$. By the continuity of the ch.f., $f_R(xt)\to f_R(0)=1$ for each $t$ as $x\downarrow0$. Thus, we have $f_C(t)=\lim_{x\in S,\,x\downarrow0}g_x(t)$, and hence $f_R(t)=g_x(t/x)/f_C(t/x)$ -- for any one $x\in S$ and all real $t$. Using these recovered ch.f.'s $f_C$ and $f_R$, it will then remain to use the inverse Fourier transform to determine the distributions of $C$ and $R$. (Of course, this does not contradict the counterexample given by Bjørn Kjos-Hanssen, since in that example $X$ takes only one positive value and the condition stated in the beginning of this answer does not hold.)

Suppose that the set $S:=\{x>0\colon P_X(x)\ne0\}$ has a nonempty intersection with each right neighborhood of $0$.
In terms of the characteristic functions (ch.f.'s) $f_C$ and $f_R$ of $C$ and $R$, your convolution equation means that $g_x(t):=f_C(t)f_R(xt)$ is known for all $x\in S$ and all real $t$. By the continuity of the ch.f., $f_R(xt)\to f_R(0)=1$ for each $t$ as $x\downarrow0$; this simple observation, that $\lim_{t\to0}f(t)=1$ for any ch.f. $f$ (together with the above condition on $S$), is crucial for the recovery of the distributions of $C$ and $R$. Indeed, now we have $f_C(t)=\lim_{x\in S,\,x\downarrow0}g_x(t)$, and hence $f_R(t)=g_x(t/x)/f_C(t/x)$ -- for any one $x\in S$ and all real $t$. Using these recovered ch.f.'s $f_C$ and $f_R$, it will then remain to use the inverse Fourier transform to determine the distributions of $C$ and $R$. (Of course, this does not contradict the counterexample given by Bjørn Kjos-Hanssen, since in that example $X$ takes only one positive value and the condition on the set $S$ stated in the beginning of this answer does not hold.)

In terms of the characteristic functions (ch.f.'s) $f_C$ and $f_R$ of $C$ and $R$, your convolution equation means that $g_x(t):=f_C(t)f_R(xt)$ is known for all real $t$ and all $x>0$ such that $P_X(x)\ne0$. If there is such an $x$ in each right neighborhood of $0$, then, letting $x\downarrow0$ and noting that then, by the continuity of the ch.f., $f_R(xt)\to f_R(0)=1$ for all $t$, we recover $f_C(t)=\lim_{x\downarrow0}g_x(t)$, and hence $f_R(t)=g_x(t/x)/f_C(t/x)$ -- for any $x>0$ such that $P_X(x)\ne0$ and all real $t$. It will then remain to use the inverse Fourier transform to determine the distributions of $C$ and $R$. (Of course, this does not contradict the counterexample given by Bjørn Kjos-Hanssen, since in that example $X$ takes only one positive value.)

Suppose that the set $S:=\{x>0\colon P_X(x)\ne0\}$ has a nonempty intersection with each right neighborhood of $0$.
In terms of the characteristic functions (ch.f.'s) $f_C$ and $f_R$ of $C$ and $R$, your convolution equation means that $g_x(t):=f_C(t)f_R(xt)$ is known for all $x\in S$ and all real $t$. By the continuity of the ch.f., $f_R(xt)\to f_R(0)=1$ for each $t$ as $x\downarrow0$. Thus, we have $f_C(t)=\lim_{x\in S,\,x\downarrow0}g_x(t)$, and hence $f_R(t)=g_x(t/x)/f_C(t/x)$ -- for any one $x\in S$ and all real $t$. Using these recovered ch.f.'s $f_C$ and $f_R$, it will then remain to use the inverse Fourier transform to determine the distributions of $C$ and $R$. (Of course, this does not contradict the counterexample given by Bjørn Kjos-Hanssen, since in that example $X$ takes only one positive value and the condition stated in the beginning of this answer does not hold.)

In terms of the characteristic functions $f_C$ and $f_R$ of $C$ and $R$, your convolution equation means that $f_C(t)f_R(xt)$ is known for all real $t$ and all $x>0$ such that $P_X(x)\ne0$. If there is such an $x$ in each right neighborhood of $0$, then, letting $x\downarrow0$ and noting that then $f_R(xt)\to1$ for all $t$, we recover $f_C$, and hence $f_R$. It will then remain to use the inverse Fourier transform. (Of course, this does not contradict the counterexample given by Bjørn Kjos-Hanssen, since in that example $X$ takes only one positive value.)

In terms of the characteristic functions (ch.f.'s) $f_C$ and $f_R$ of $C$ and $R$, your convolution equation means that $g_x(t):=f_C(t)f_R(xt)$ is known for all real $t$ and all $x>0$ such that $P_X(x)\ne0$. If there is such an $x$ in each right neighborhood of $0$, then, letting $x\downarrow0$ and noting that then, by the continuity of the ch.f., $f_R(xt)\to f_R(0)=1$ for all $t$, we recover $f_C(t)=\lim_{x\downarrow0}g_x(t)$, and hence $f_R(t)=g_x(t/x)/f_C(t/x)$ -- for any $x>0$ such that $P_X(x)\ne0$ and all real $t$. It will then remain to use the inverse Fourier transform to determine the distributions of $C$ and $R$. (Of course, this does not contradict the counterexample given by Bjørn Kjos-Hanssen, since in that example $X$ takes only one positive value.)