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A recipe for a counterexample:

Let $(X,e,m:X\wedge X\to X)$ be a monoid object in the homotopy category of pointed spaces $h\mathcal{S}_\ast$ (that is, an H-monoid), where $\wedge$ denotes the smash product in the pointed homotopy category. Then we can form an $h\mathcal{S}$-enriched category $\mathbf{B}X$ with set of objects $\{\ast\}$ and such that $\mathbf{B}X(\ast,\ast)=X,$ with composition given by $$\mu:X\times X \to X\wedge X \xrightarrow{m} X.$$ and unit given by $e:\Delta^0 \to X$ the basepoint of $X$.

In general, an H-monoid structure $(X,e,m)$ only specifies a canonical $A_2$-algebra structure on $X$ with the added stipulation that there exists an $A_3$-structure extending the $A_2$-structure, and there exist many examples (none of which I know!) of $A_2$-structures admitting (possibly many) extensions to an $A_3$ structure, none of which extend to an $A_\infty$-structure. Unfortunately, even having asked some experts, it's not so easy to come up with an example of such a space.

But it is a theorem that every $A_\infty$-space is equivalent to an honest monoid in spaces.

So it suffices to find any H-monoid $(X,e,m)$ in spaces that doesn't admit a lift to an $A_\infty$-monoid, then take $\mathbf{B}X$. This gives an example of an $h\mathcal{S}$-enriched category that cannot lift to an honest $\mathcal{S}$-enriched category, because if it did, it would necessarily specify an $A_\infty$-structure on $X$ lifting $(X,e,m)$.

A recipe for a counterexample:

Let $(X,e:\Delta^0\to X,m:X\times X\to X)$ be monoid object in the homotopy category of spaces $h\mathcal{S}$ (that is, an H-monoid). Note that this is a property of the triple. Then we can form an $h\mathcal{S}$-enriched category $\mathbf{B}X$ with set of objects $\{\ast\}$ and such that $\mathbf{B}X(\ast,\ast)=X,$ with composition given by $$m:X\times X \to X.$$ and unit given by $e:\Delta^0 \to X$ the basepoint of $X$.

In general, an H-monoid structure $(X,e,m)$ only specifies a canonical $A_2$-algebra structure on $X$ with the added stipulation that there exists an $A_3$-structure extending the $A_2$-structure, and there exist many examples (none of which I know!) of $A_2$-structures admitting (possibly many) extensions to an $A_3$ structure, none of which extend to an $A_\infty$-structure. Unfortunately, even having asked some experts, it's not so easy to come up with an example of such a space.

But it is a theorem that every $A_\infty$-space is equivalent to an honest monoid in spaces.

So it suffices to find any H-monoid $(X,e,m)$ in spaces that doesn't admit a lift to an $A_\infty$-monoid, then take $\mathbf{B}X$. This gives an example of an $h\mathcal{S}$-enriched category that cannot lift to an honest $\mathcal{S}$-enriched category, because if it did, it would necessarily specify an $A_\infty$-structure on $X$ lifting $(X,e,m)$.

Edit: Kevin Carlson's comment below gives an example of such spaces from Stasheff and Adams and a source, so check it out!

A recipe for a counterexample:

Let $(X,e,m:X\wedge X\to X)$ be a monoid object in the homotopy category of pointed spaces $h\mathcal{S}_\ast$ (that is, an H-monoid), where $\wedge$ denotes the smash product in the pointed homotopy category. Then we can form an $h\mathcal{S}$-enriched category $\mathbf{B}X$ with set of objects $\{\ast\}$ and such that $\mathbf{B}X(\ast,\ast)=X,$ with composition given by $$\mu:X\times X \to X\wedge X \xrightarrow{m} X.$$ and unit given by $e:\Delta^0 \to X$ the basepoint of $X$.

In general, an H-monoid structure $(X,e,m)$ only specifies a canonical $A_3$-algebra structure on $X$, and there exist many examples (none of which I know!) of $A_3$-structures that don't admit a lift to an $A_\infty$ structure.

But it is a theorem that every $A_\infty$-space is equivalent to an honest monoid in spaces.

So it suffices to find any H-monoid $(X,e,m)$ in spaces that doesn't admit a lift to an $A_\infty$-monoid, then take $\mathbf{B}X$. This gives an example of an $h\mathcal{S}$-enriched category that cannot lift to an honest $\mathcal{S}$-enriched category, because if it did, it would necessarily specify an $A_\infty$-structure on $X$ lifting $(X,e,m)$.

A recipe for a counterexample:

Let $(X,e,m:X\wedge X\to X)$ be a monoid object in the homotopy category of pointed spaces $h\mathcal{S}_\ast$ (that is, an H-monoid), where $\wedge$ denotes the smash product in the pointed homotopy category. Then we can form an $h\mathcal{S}$-enriched category $\mathbf{B}X$ with set of objects $\{\ast\}$ and such that $\mathbf{B}X(\ast,\ast)=X,$ with composition given by $$\mu:X\times X \to X\wedge X \xrightarrow{m} X.$$ and unit given by $e:\Delta^0 \to X$ the basepoint of $X$.

In general, an H-monoid structure $(X,e,m)$ only specifies a canonical $A_2$-algebra structure on $X$ with the added stipulation that there exists an $A_3$-structure extending the $A_2$-structure, and there exist many examples (none of which I know!) of $A_2$-structures admitting (possibly many) extensions to an $A_3$ structure, none of which extend to an $A_\infty$-structure. Unfortunately, even having asked some experts, it's not so easy to come up with an example of such a space.

But it is a theorem that every $A_\infty$-space is equivalent to an honest monoid in spaces.

So it suffices to find any H-monoid $(X,e,m)$ in spaces that doesn't admit a lift to an $A_\infty$-monoid, then take $\mathbf{B}X$. This gives an example of an $h\mathcal{S}$-enriched category that cannot lift to an honest $\mathcal{S}$-enriched category, because if it did, it would necessarily specify an $A_\infty$-structure on $X$ lifting $(X,e,m)$.

Edit: I don't think $S^7$ is associative actually, but the same argument works with any H-monoid that doesn't lift to an $E_1$-monoid

Original answer:

An explicit counterexample: Consider the 7-sphere with its H-space structure in the homotopy category of spaces. Then consider $BS^7$ to be the one-object $h\mathcal{S}_\ast$-enriched category whose unique endomorphism object and compositions are obtained from the H-space structure on $S^7$. Then an argument [1] shows that $BS^7$ cannot arise as the homotopy category of an $\mathcal{S}_\ast$-enriched category, since such a structure must arise from an $E_1$ structure.

[1] https://ncatlab.org/nlab/show/H-space#spheres

A recipe for a counterexample:

Let $(X,e,m:X\wedge X\to X)$ be a monoid object in the homotopy category of pointed spaces $h\mathcal{S}_\ast$ (that is, an H-monoid), where $\wedge$ denotes the smash product in the pointed homotopy category. Then we can form an $h\mathcal{S}$-enriched category $\mathbf{B}X$ with set of objects $\{\ast\}$ and such that $\mathbf{B}X(\ast,\ast)=X,$ with composition given by $$\mu:X\times X \to X\wedge X \xrightarrow{m} X.$$ and unit given by $e:\Delta^0 \to X$ the basepoint of $X$.

In general, an H-monoid structure $(X,e,m)$ only specifies a canonical $A_3$-algebra structure on $X$, and there exist many examples (none of which I know!) of $A_3$-structures that don't admit a lift to an $A_\infty$ structure.

But it is a theorem that every $A_\infty$-space is equivalent to an honest monoid in spaces.

So it suffices to find any H-monoid $(X,e,m)$ in spaces that doesn't admit a lift to an $A_\infty$-monoid, then take $\mathbf{B}X$. This gives an example of an $h\mathcal{S}$-enriched category that cannot lift to an honest $\mathcal{S}$-enriched category, because if it did, it would necessarily specify an $A_\infty$-structure on $X$ lifting $(X,e,m)$.