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 $\mathbf{S}_{X/}$.
Edit: Actually the monoidal product on $\mathbf{S}_{X/}$ cannot not be taken to be the Cartesian one, because the Cartesian product on $\mathbf{S}_{X/}$ doesn't preserve colimits in each variable separately, so this argument doesn't work as is. However, all the rest seems to still be correct, so that $\mathbf{S}_{X/}$ will be an excellent model category as soon as it has some compatible monoidal structure. In particular, $\mathbf{S}_{X/}$ satisfies (A2) but not every object is cofibrant in general.
Second Edit: Constructing monoidal structures on $\mathbf{S}_{X/}$ turned out to be quite of a headache. However, there is a natural left action of $\mathbf{S}$ on $\mathbf{S}_{X/}$ which is given by
$$ Z \otimes (X \to Y) := X \to X \coprod_{Z \times X} \left[Z \times Y\right] $$
and a natural right action given by
$$ (X \to Y) \otimes Z := X \to X \coprod_{X \times Z} \left[Y \times Z\right] .$$
Here the pushouts are performed with respect to the projections $Z \times X \to X$ and $X \times Z \to X$ respectively (these two actions are naturally isomorphic, but it is convenient to have them written on both sides). These actions preserve colimits in each variable separately, and it is not hard to verify that if the Cartesian product on $\mathbf{S}$ satisfies the pushout-product axiom then these actions satisfy the pushout-product axiom as well. We may hence think of $\mathbf{S}_{X/}$ as a bimodule model category over $\mathbf{S}$. In this case we can construct a new (non-Cartesian) monoidal model category, the "square-zero extension" $\mathbf{S}_{X/} \rtimes \mathbf{S}$, whose underlying model category is the product model category $\mathbf{S}_{X/} \times \mathbf{S}$, and such that
$$(Z,X \to Y) \otimes (Z',X' \to Y') := (Z \times Z',\left[Z \otimes (X' \to Y')\right] \coprod \left[(X \to Y) \otimes Z'\right])$$
When $\mathbf{S}$ is an excellent Cartesian model category (e.g. simplicial sets) then $\mathbf{S}_{/X}$ satisfies all the axioms of an excellent model category except the existence of a monoidal structure (see above). This implies that the monoidal model category $\mathbf{S}_{X/} \rtimes \mathbf{S}$ is actually an excellent model category. However, in general not every object in $\mathbf{S}_{X/} \rtimes \mathbf{S}$ is cofibrant, e.g. when $\mathbf{S}$ is simplicial sets and $X = \Delta^0 \coprod \Delta^0$.
The conclusion of this discussion is not that provocative. Since the condition that every object is cofibrant is required (or at least used) for setting up a model structure on $\mathbf{S}$-enriched categories (see Proposition A.3.2.4 of HTT), all this means is that one should simply add "every object is cofibrant" to the list of axioms defining an excellent model category.
