This is not an answer, but too long for a comment. Below I write down some conditions (on a group or a decomposable set) guaranteeing that a decomposable set in a group is product-one.

**Proposition 1.** Let $G$ be a group containing an abelian subgroup $A$ of index 2 such that for every $x\in A$ and $y\in G\setminus A$ we have $yx=x^{-1}y$. Every finite decomposable set $D$ in $G$ is product-one.

*Proof.* First observe that the intersection $D\cap A$ is not empty. Otherwise, $D\subseteq DD=(D\setminus A)\cdot(D\setminus A)\subseteq (G\setminus A)(G\setminus A)=A$ would be empty. If $D\cap A$ contains a decomposable subset of the abelian group $A$, then it is product-one by the result of Lev, Nagy, and Pach. So, we assume that $D\cap A$ contains no decomposable subset. In particular, $D$ does not contain the identity 1 of the group $G$. Since $D\cap A$ is not decomposable, there exists an element $x\in D\cap A$ such that $x=yz$ for some elements $y,z\in D\setminus A$. It follows from $1\notin D$ that $y,z$ differ from $x$.

First we assume that $y\ne z$. Then $yxz=x^{-1}yz=x^{-1}x=1$, witnessing that $D$ is product-one.

Now assume that $y=z$. In this case $x=yz=y^2$. Since $D$ is decomposable, $y=uv$ for some elements $u,v\in D$. Since $1\notin D$, the elements $u,v$ differ from $y$. Assuming that $u=v$, we conclude that $y=uv=u^2\in A$ as the group $A$ has index 2 in $G$. But this contradicts the choice of the points $y=z\notin A$. So, $u\ne v$. If $x\notin\{u,v\}$, then $yxuv=x^{-1}yuv=x^{-1}y^2=x^{-1}x=1$ and we are done. So, we assume that $x\in\{u,v\}$. If $x=u$, then $x=y^2=uvy=xvy$ implies $vy=1$. If $x=v$, then $x=y^2=yuv=yux$ implies $yu=1$. In both cases we have found two elements in $D$ whose product is equal 1, witnessing that $D$ is product-one.

**Corollary.** Decomposable sets in the dihedral groups $D_{2n}$ and dicyclic groups $Q_{4n}$ are product-one.

**Proposition 2.** If a finite decomposable subset $D$ of a group contains an element of order 2, then $D$ is product-one.

*Proof.* If $D$ contains the identity of the group, then $D$ is product-one. So, we assume that $1\notin D$. If $D$ contains a sequence of pairwise distinct points $x_0,\dots,x_n$ such that $x_{k-1}=x_k^2$ for any positive $k<n$ and $x_n=x_0^2$, then $$x_0=x_1^2=x_1x_2^2=\dots =x_1x_2\cdots x_{n-1}x_n^2=x_1x_2\cdots x_nx_0^2$$which implies that $x_1x_2\cdots x_nx_0=1$ and hence $D$ is product-one. So, we assume that $D$ does not contain such a sequence. Fix any element $x\in D$ of order 2 and let $n\in\mathbb N$ be the largest number for which there exists a sequence $x_0,\dots,x_n$ in $D$ such that $x_0=x$ and $x_{k-1}=x_k^2$ for all positive $k\le n$. Our assumption guarantees that $n$ is well-defined and the points $x_0,\dots,x_n$ are pairwise distinct. By the decomposability of $D$, there exist two element $y,z\in D$ such that $x_n=yz$. The maximality of $n$ guarantees that $y\ne z$. If the doubleton $\{y,z\}$ is disjoint with the set $\{x_0,\dots,x_n\}$, then $$1=x^2=x_0x_1\cdots x_nyz$$and hence $D$ is product-one. So we assume that $\{y,z\}\cap\{x_0,\dots,x_n\}\ne\emptyset$ and find the largest number $i$ such that $x_i\in\{y,z\}$. It follows from $x_n=yz$ and $1\notin D$ that $i<n$. If $x_i=z$, then $$x_i=x_{i+1}^2=x_{i+1}\cdots x_nyz=x_{i+1}\cdots x_nyx_i$$implies that $x_{i+1}\cdots x_ny=1$, witnessing that $D$ is product-one. If $x_i=y$, then
$$x_i=x_{i+1}^2=yzx_n\cdots x_{i+1}=x_izx_n\cdots x_{i+1}$$ implies that $zx_n\cdots x_{i+1}=1$, which means that $D$ is product-one.

**Proposition 3.** Every decomposable set $D$ of cardinality $|D|\le 3$ is product-one.

*Proof.* If $D=\{a\}$ is a singleton, then $a=a^2$ and $a=1$, which means that $D$ is product-one. Now assume that $D=\{a,b\}$ is a doubleton and $1\notin D$. Then $a=b^2$ and $b=a^2=b^4$, which implies that $ab=b^3=1$ and hence $D$ is product-one. Finally, assume that $D=\{a,b,c\}$ and $D$ contains no decomposable subsets of cardinality $<3$.

First we assume that some element of $D$ is the square of some other element of $D$. We lose no generality assuming that $b=a^2$. By the decomposability of $D$, $c\in\{a^2,b^2,ab,ba\}=\{a^2,a^4,a^3\}$ and hence $\{a,b,c\}$ is a decomposable subset of the abelian group $\{a^n:n\in\mathbb Z\}$. By the result of Lev, Nagy and Pach, the decomposable set $D$ is product-one.

If $D$ contains an element of order 2, then $D$ is product-open by Proposition 2.

It remains to consider the case when no element of $D$ is the square of another element of $D$ and no element of $D$ has order 2. By the decomposability of $D$, $a=bc$ or $a=cb$. We lose no generality assuming that $a=bc$ and hence $c=b^{-1}a$. By the decomposability of $D$, $b=ac$ or $b=ca$. If $b=ac$, then $b=ac=ab^{-1}a$ and hence $c^2=(b^{-1}a)^2=1$, which is forbidden by our assumption. So, $b=ca=b^{-1}a^2$ and hence $b^2=a^2$.

By the decomposability of $D$, $c=ab$ or $c=ba$. If $c=ba$, then $ba=c=b^{-1}a$ and hence $b^2=1$, which is forbidden by our assumption. So, $ab=c=b^{-1}a$ and hence $aba^{-1}=b^{-1}$. Now consider the group $H$ generated by the elements $a,b$ and the cyclic subgroup $B$ generated by the element $b$. It follows from $a^2=b^2\in B$ and $aba^{-1}=b^{-1}\ne b$ that the group $B$ has index 2 in $H$ and for every elements $x=b^n\in B$ and $y=ab^m\in H\setminus B$ we have $yx=ab^mb^n=ab^nb^m=b^{-n}ab^m=x^{-1}y$. Since $D=\{a,b,c\}=\{a,b,b^{-1}a\}\subseteq H$, we can apply Proposition 1 and conclude that the decomposable set $D$ is product-one.

3more comments