Suppose that $h(Y)=U+Y$, and $Y,U$ are independent. Then $U=Z:=h(Y)-Y$ and $U,Z$ are independent. So, $U$ is independent of itself. So, $U$ is constant almost surely (a.s.): $P(U=u)=1$ for some real $u$. 

Vice versa, if $P(U=u)=1$ for some real $u$, then $h(Y)=U+Y$ a.s. with $h(Y):=u+Y$. 

Thus, if $Y,U$ are independent, then there exists a (Borel-measyrable) function $h$ such that $h(Y)=U+Y$ if and only if $U$ is constant a.s.