Example of a $U^\omega$ that is not Deterministic Büchi recognizable
Take an alphabet $A=\{0,1,2\}$, and $U=0^*+2A^*$. Then, $U^\omega$ is not deterministic Büchi recognizable.

To see this, imagine it is recognized by a deterministic Büchi automaton $\mathcal A$ with $n$ states. Then, look at the run of $\mathcal A$ on the word $w=2(0^{n+1}1)^\omega$.
Any inside loop of $0$ must contain a Büchi state, because otherwise pumping yields a contradiction (we get a word of the form $2A^*0^\omega$ that is in $U^\omega$ but not accepted by $\mathcal A$). But then, it means that $w$ is accepted, and therefore $\mathcal A$ does not recognize exactly $U^\omega$.