No, this is not inconsistent, at least not relative to the existence of a Mahlo cardinal.
Suppose that $\kappa$ is a Mahlo cardinal, then $V_\kappa$ has a club of ordinals $\alpha$ such that $V_\alpha\prec V_\kappa$. Therefore there is a stationary set of inaccessible cardinals satisfying this.
Simply enumerate these inaccessible cardinals and let $\alpha$ be the enumeration of this stationary set. Then $U_\alpha=V_\mu$ for some inaccessible $\mu$ and $V_\mu\prec V_\kappa$.
Now cut the universe at $V_\kappa$, and this satisfies this extension of $\sf ZFC$ that you suggest. You can probably get away with much less than a Mahlo.