I assume that $U^\ast_{ij}=S(u_j^i)$, where $S$ is the antipode. (See for example the book by Klimyk & Schmüdgen, Section 9.2.4.) If so, then $UU^\ast=1=U^\ast U$ follows from the antipode axiom

$\mu\circ(\mathrm{Id}\otimes S)\circ \Delta =\varepsilon =\mu\circ( S\otimes\mathrm{Id})\circ \Delta$

after applying both sides to an arbitrary generator $u_j^i$.

To prove $u_j^iS(u_s^r)=S(u_s^r)u_j^i$ for $r\neq i, s\neq j$, one can multiply both sides by the quantum determinant and use that it is central to get the equivalent statement $u_j^iM_s^r=M_s^r u_j^i$, where $M_s^r$ are quantum minors of size $(N-1)\times(N-1)$.
Now one can observe that the subalgebra generated by $u_b^a$ with $a\neq r, b\neq s$ is isomorphic to $M_q(N-1,C)$, the quantized algebra of regular functions on all $(N-1)\times(N-1)$ complex matrices. $u_j^i$ and $M_s^r$ belong to this subalgebra and moreover $M_s^r$ is the quantum determinant in $M_q(N-1,C)$, hence it commutes with $u_j^i$.