I am now under the impression that it is simply not true that in an excellent model category every object is cofibrant. Let $\mathbf{S}$ be an excellent model category in which the monoidal structure is the Cartesian one (e.g., simplicial sets). For every $X \in \mathbf{S}$ we may consider the coslice category $\mathbf{S}_{X/}$ with its canonical model structure, in which all three classes are created by the projection $p:\mathbf{S}_{X/} \to \mathbf{S}$. We now note that $p$ creates limits and connected colimits (i.e., colimits indexed by connected diagrams), and in particular pushouts. It then follows that the Cartesian product on $\mathbf{S}_{X/}$ is compatible with the model structure. We now claim that $\mathbf{S}_{X/}$ is an excellent model category (with respect to the Cartesian monoidal structure). First note that $\mathbf{S}_{X/}$ is combinatorial. Property (A2) discussed above follows from the same property for $\mathbf{S}$ since $p:\mathbf{S}_{X/} \to \mathbf{S}$ is a right functor and hence preserves products and monomorphisms (while creating cofibrations by definition). Weak equivalences in $\mathbf{S}_{X/}$ are closed under filtered colimits since $p$ preserves filtered colimits and detects weak equivalences. Finally, the invertibility hypothesis follows from Lemma A.3.2.20 of HTT since we have a left Quillen functor $\mathbf{S} \to \mathbf{S}_{X/}$.

On the other hand, $\mathbf{S}_{X/}$ will typically **not** have all its objects cofibrant. Take for example $\mathbf{S}$ to be simplicial sets and $X = \Delta^0 \coprod \Delta^0$, in which case the object $X \to \Delta^0$ is not cofibrant in $\mathbf{S}_{X/}$.