Taking $\rho_{AB}=\rho_{A}\otimes \rho_{B}$, where $S(\rho_{A})$ and $S(\rho_{B})$ aren't 0, it's easy to see that
$$S(\rho_{AB}||I \otimes \rho_{B})=-S(\rho_{A})-S(\rho_{B})+S(\rho_{B})=-S(\rho_{A}).$$
Now clearly this is less than 0 due to the non-negativtiy of the von-neumann entropy. However, the non-negativty of the relative entropy is given in every textbook, with the caveat that for $S(\rho||\sigma) \ge0$ provided the support of $\rho\subseteq \sigma$, which by the violation of the inequality, can't be the case in this example. Although I don't see how, as if I take each marginal to be maximally mixed, the above still holds, yet in that case the support of $\rho \subseteq \sigma$. Perhaps I am missing something here?
However, if $\rho_{AB}$ were a maximally entangled state, it would be non-negative. So given cases like this, wherein the support isn't a subset of the second argument, are there other ways to show it will hold? Possibly something to do with $S(\rho_{AB}||N(\rho_{AB}))$ wherein $N$ is some CPTP channel?