I can only answer half of your question: namely, a standard condition under which the more general unit axiom holds. I don't know of any examples where Hovey's unit axiom holds but this more general one does not. The hypothesis is that **cofibrant objects are flat**, i.e. smashing with cofibrant objects preserves weak equivalences. This hypothesis comes up all the time in Hovey's work, and also appears in the paper of Schwede-Shipley where the Monoid Axiom is first introduced. There you need cofibrant objects to be flat in order to conclude that if $R\rightarrow S$ is a weak equivalence of ring objects, then $Ho(R-mod)\cong Ho(S-mod)$.

Suppose that cofibrant objects are flat. Then we know Hovey's unit axiom automatically, since $QI\rightarrow I$ is a weak equivalence and so for any cofibrant $Y$, $Y$ smashed with this map is still a weak equivalence. To see that the more general unit axiom holds, let $X$ be any object (not necessarily cofibrant). Then we have the following commutative diagram:

$$ \begin{array}{rrrr}
QI\otimes QX & \rightarrow & QX & \rightarrow & X\\\
\downarrow & & & & \downarrow \\\
QI \otimes X & & \rightarrow & & X
\end{array}
$$

Here the top maps are weak equivalences by Hovey's unit axiom and by the definition of $QX$. The left vertical map is a weak equivalence because cofibrant objects are flat and $QI$ is cofibrant. The right vertical is clearly a weak equivalence because it's the identity. Thus, by the 2-out-of-3 property, the bottom horizontal is a weak equivalence.

EDIT: It should go without saying, but in a non-symmetric setting "cofibrant objects are flat" means both $K\otimes f$ and $f\otimes K$ are weak equivalences. So the argument above works if you apply the twist natural transformation to everything, and we also have $X\otimes QI \rightarrow X$ is a weak equivalence.