From what I understand, the localization of a ring $R$ with a multiplicative set $S$ is a construction of a bigger ring $R^*$ such that $R$ is a subring in this bigger ring and every element in $S$ is a unit.
Furthermore, we want that if we have any ring homomorphism $\phi$ from $R$ to another ring $T$ such that $\phi(d)$ is a unit ($d \in R$) and $\phi(1) = 1$, we also want that the construction of $R^*$ satisfies the universal property. So there is a unique homomorphism $\pi$ from $R^*$ to $T$ such that $\phi = \pi \circ \rho$ where $\rho$ is the homomorphism from $R$ to $R^*$.
Now my question is (sorry it takes so long to get to the point) what if I had that my multiplicative set contains two zero divisors $d$ and $d'$ where $dd'=0$? How can any ring homomorphism map any zero divisors to a unit element in $T$?
Since $\phi(dd') = \phi(0) = 0 = \phi(d)\phi(d')$.
So $\phi(d)$ and $\phi(d')$ are zero divisors, hence they cannot be units.
Where did my reasoning go wrong?