This is a follow-up question to Is there a non-free group $G$ whose subgroups are all freely decomposable?

In the answer to that question, Cornulier gives the following example (due to Kurosh) of a group $G$ which is not free, yet isomorphic to $G*\mathbb{Z}$:

$$G=\langle (a_n)_{n\geq 0},(b_n)_{n\geq 1},\mid \left[a_n,b_n\right]=a_{n-1},\forall n\geq 1\rangle$$

Of course $G\cong G*F$ whenever $F$ is a finitely generated free group.

Question: Is the group

$$H=\langle (a_n)_{n\geq 0},(b_n)_{n\geq 1},(c_n)_{n\geq 1},(d_n)_{n\geq 1}\mid \left[a_n,b_n\right]\left[c_n,d_n\right]=a_{n-1},\forall n\geq 1\rangle$$

non-isomorphic to $H*F_n$, $n=1,2\;(\operatorname{mod}3)$? ($H$ is isomorphic to $H*F_3$.) Is $H$ isomorphic to $G$? It seems to me like it shouldn't be.