Take $R$ to be the coordinate ring of the nodal curve $\mathbb{C}[t^2-1, t(t^2-1)]$ and $S$ to be its normalization $\mathbb{C}[t]$. It satisfies (1), ..., (4): The first three are immediate. For (4), note that since $R$ is noetherian, the projectivity of $S$ as an $S \otimes_R S$-module is equivalent to $S$ being unramified over $R$ (Theorem 2.5 in Auslander-Buchsbaum, _On Ramification Theory in Noetherian Rings_, American Journal of Mathematics, 1959). It is enough to check that maximal ideals are unramified. Since we are in equi-characteristic zero, it suffices to show that for every maximal ideal $\mathfrak{q}$ of $S$, $(\mathfrak{q} \cap R) S_{\mathfrak{q}} = \mathfrak{q} S_{\mathfrak{q}}$. Let $\mathfrak{q} = (t-\alpha)$, $\alpha \in \mathbb{C}$.
Then $(\mathfrak{q} \cap R) S = 
(t^2 - \alpha^2, t(t^2-1)- \alpha(\alpha^2-1))$.
If $\alpha \neq 0$, then $t+\alpha \not \in \mathfrak{q}$
so $t-\alpha \in (\mathfrak{q} \cap R)S_{\mathfrak{q}}$.
If $\alpha = 0$, then $t^2-1 \not \in \mathfrak{q}$
so $t \in (\mathfrak{q} \cap R)S_{\mathfrak{q}}$.
Either way, $(\mathfrak{q} \cap R)S_{\mathfrak{q}} = 
\mathfrak{q}S_{\mathfrak{q}}$. However, $S$ is not a flat $R$-module.