The answer is no. E.g., let $U=(0,1/2)$ and $$\zeta=2\times 1_{(0,1/2)^2}+2\times 1_{(1/2,1)^2};$$ that is, the joint distribution of $X,Y)$ is the half-and-half mixture of the uniform distributions on the the squares $(0,1/2)^2$ and $(1/2,1)^2$. Then $$P_{X,Y}(U\times U)=\frac12\ne\frac12\times\frac12=P_X(U)P_Y(U).$$

The answer is no. E.g., let $U=(0,1)$ and $$\zeta=\frac12\times 1_{(0,1)^2}+\frac12\times 1_{(1,2)^2};$$ that is, the joint distribution of $(X,Y)$ is the half-and-half mixture of the uniform distributions on the the squares $(0,1)^2$ and $(1,2)^2$. Then $$P_{X,Y}(U\times U)=\frac12\ne\frac12\times\frac12=P_X(U)P_Y(U).$$

The answer is no. E.g., let $U=(0,1/2)$ and $$\zeta(x,y)=2\times1(0<x,y<1/2)+2\times1(1/2<x,y<1)$$ for all real $x,y$; that is, the joint distribution of $X,Y)$ is the half-and-half mixture of the uniform distributions on the the squares $(0,1/2)^2$ and $(1/2,1)^2$. Then $$P_{X,Y}(U\times U)=\frac12\ne\frac12\times\frac12=P_X(U)P_Y(U).$$

The answer is no. E.g., let $U=(0,1/2)$ and $$\zeta=2\times 1_{(0,1/2)^2}+2\times 1_{(1/2,1)^2};$$ that is, the joint distribution of $X,Y)$ is the half-and-half mixture of the uniform distributions on the the squares $(0,1/2)^2$ and $(1/2,1)^2$. Then $$P_{X,Y}(U\times U)=\frac12\ne\frac12\times\frac12=P_X(U)P_Y(U).$$