I can only answer half of your question: namely, a standard condition under which your unit axiom holds. I don't know of any examples where Hovey's unit axiom holds but yours 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 we know Hovey's unit axiom and that cofibrant objects are flat. Let $X$ be any object. 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.

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.

I can only answer half of your question: namely, a standard condition under which your unit axiom holds. I don't know of any examples where Hovey's unit axiom holds but yours 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 we know Hovey's unit axiom and that cofibrant objects are flat. Let $X$ be any object. 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.

I can only answer half of your question: namely, a standard condition under which your unit axiom holds. I don't know of any examples where Hovey's unit axiom holds but yours 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 we know Hovey's unit axiom and that cofibrant objects are flat. Let $X$ be any object. 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.