Thus $G$ gives an $A_3$-monoidal groupoid which does not arise from an $A_4$-monoidal one, equivalently, not from a strictly monoidal one.  EDIT: As Harry notes, this is not quite a counterexample to your question, as it's not clear that it's impossible to make a different choice of the associators that would extend to $A_4$. It seems unlikely, but also that it would be complicated to make an elementary argument against it.

Thus $G$ gives an $A_3$-monoidal groupoid which does not arise from an $A_4$-monoidal one, equivalently, not from a strictly monoidal one. 

EDIT: As Harry notes, this is not quite a counterexample to your question, as it's not clear that it's impossible to make a different choice of the associators that would extend to $A_4$. It seems unlikely, but also that it would be complicated to make an elementary argument against it.

Here is a proposed easier example, coming from $\mathcal C=\mathrm{Cat}$ instead of $\mathrm{Gpd}$, again with the canonical model structure. We let $C$ be the free $A_3$-monoidal category generated by a co-pointed object. So as compared to $G$ from above, $C$ contains a map $p:x\to I$ contracting a leaf which freely generates $C$ from $G$ under $\otimes$ and naturality with respect to the elementary moves. I'll refer to any tensor product of $p$'s and identities as a projection. Call the canonical associators, given by elementary moves, $\alpha$; I claim there is no alternative choice $\beta$ of associators for an $A_3$-structure on $C$, so that $C$'s underlying $\mathrm{Ho}(\mathrm{Cat})$-enriched category arises from no monoidal category.

To show that $C$ admits no $A_4$-monoidal structure, note that $\beta_{x,x,x}:(xx)x\to x(xx)$ is uniquely determined as it was in $G$-we've added no new morphisms between isomorphic trees. Furthermore, while morphisms between trees with different numbers of leaves cannot be uniquely written as strings of elementary moves and projections, the only relation on them is naturality of projections with respect to elementary moves. The induced equivalence relation respects lengths of strings of elementary moves and projections, so that any morphism of $C$ has a well defined length given by the length of any representing string of elementary moves and projections.

Then to show for instance that $\beta_{xx,x,x}=\alpha_{xx,x,x}$, we can consider the equality $$(px)(xx)\circ \beta_{xx,x,x}=\beta_{x,x,x}\circ ((px)x)x:((xx)x)x\to x(xx).$$ This implies that $\beta_{xx,x,x}$ must be an elementary move, by computing lengths. And there is a unique elementary move $((xx)x)x\to (xx)(xx)$, namely $\alpha_{xx,x,x}$. Similarly one shows all the associators between four-leaf trees are the canonical ones, so that the pentagon cannot commute.

If I haven't missed anything in this example, it seems as if its nerve should give rise to an $A_3$-but-not-$A_4$ space. I can't tell if it can be made to give such a groupoid, though.

Summarizing a discussion in the comments on Harry's answer, we can consider the case $\mathcal C=\mathrm{Gpd}$, with the canonical model structure: weak equivalences are just the equivalences of groupoids, cofibrations are functors injective on objects, and fibrations are isofibrations.

A $\mathcal C$-enriched category with one object is then precisely a strictly monoidal groupoid. In contrast, a $\mathrm{Ho}(\mathcal C)$-enriched category with one object is essentially the same thing as an $A_3$-monoidal groupoid-we get a tensor product, which is certainly associative up to non-specified isomorphism, but there is no reason why the associators should satisfy the pentagon identity, and indeed they need not.

For a counterexample, consider the $A_3$-monoidal groupoid $G$ freely generated by a single object $x$. For simplicity, let's also make it strictly unital. Then the objects of $G$ are given by rooted binary trees, corresponding to parenthesizations of finite strings of $x$'es. The morphisms are freely generated by the associator isomorphism between the two rooted binary trees of three leaves, together with the functoriality of the tensor product $\otimes:G\times G\to G$ which fuses two trees by adding a new root to their disjoint union.

It is in fact possible to describe the morphisms in $G$ without reference to the tensor product: they are sequences of elementary moves, where an elementary move out of a binary tree $T$ is given by taking a node $N$ which is the left child of its parent $P$ and replacing $P$'s right child with $N$, $P$'s left child with $N$'s left child, $N$'s left child with its right child, and $N$'s right child with $P$'s right child. Under this description, one can see that the two different paths from $((xx)x)x$ to $x(x(xx))$ around the pentagon are not equal in $G$. (For comparison, the free strictly unital $A_4$-monoidal groupoid on $x$ has exactly one morphism between two binary trees whenever there exists any such sequence of elementary moves transforming one into the other.)

Thus $G$ gives an $A_3$-monoidal groupoid which does not arise from an $A_4$-monoidal one, equivalently, not from a strictly monoidal one.

Summarizing a discussion in the comments on Harry's answer, we can consider the case $\mathcal C=\mathrm{Gpd}$, with the canonical model structure: weak equivalences are just the equivalences of groupoids, cofibrations are functors injective on objects, and fibrations are isofibrations.

A $\mathcal C$-enriched category with one object is then precisely a strictly monoidal groupoid. In contrast, a $\mathrm{Ho}(\mathcal C)$-enriched category with one object is essentially the same thing as an $A_3$-monoidal groupoid-we get a tensor product, which is certainly associative up to non-specified isomorphism, but there is no reason why the associators should satisfy the pentagon identity, and indeed they need not.

For a counterexample, consider the $A_3$-monoidal groupoid $G$ freely generated by a single object $x$. For simplicity, let's also make it strictly unital. Then the objects of $G$ are given by rooted binary trees, corresponding to parenthesizations of finite strings of $x$'es. The morphisms are freely generated by the associator isomorphism between the two rooted binary trees of three leaves, together with the functoriality of the tensor product $\otimes:G\times G\to G$ which fuses two trees by adding a new root to their disjoint union.

It is in fact possible to describe the morphisms in $G$ without reference to the tensor product: they are sequences of elementary moves, where an elementary move out of a binary tree $T$ is given by taking a node $N$ which is the left child of its parent $P$ and replacing $P$'s right child with $N$, $P$'s left child with $N$'s left child, $N$'s left child with its right child, and $N$'s right child with $P$'s right child. Under this description, one can see that the two different paths from $((xx)x)x$ to $x(x(xx))$ around the pentagon are not equal in $G$. (For comparison, the free strictly unital $A_4$-monoidal groupoid on $x$ has exactly one morphism between two binary trees whenever there exists any such sequence of elementary moves transforming one into the other.)

Thus $G$ gives an $A_3$-monoidal groupoid which does not arise from an $A_4$-monoidal one, equivalently, not from a strictly monoidal one. EDIT: As Harry notes, this is not quite a counterexample to your question, as it's not clear that it's impossible to make a different choice of the associators that would extend to $A_4$. It seems unlikely, but also that it would be complicated to make an elementary argument against it.