There exists an infinite cardinal $\zeta$ such that there exists a set $P$ such that $P$ is a partition on $\mathcal P(\zeta) \setminus \{\varnothing\}$ such that each element $h$ of $P$ is an infinite set of bijective subsets of $\zeta$, and if an element of $h$ is finite then then $h$ is an equivalence class of subsets of $\zeta$ under bijection. And for each elements $k,l$ of $P$ if we have $k_1,k_2 \in k$ and $l_1,l_2 \in l$ such that $l_1 \subsetneq k_1,l_2 \subsetneq k_2$, then the sets $k_1 - l_1; k_2-l_2$ are elements of the same element of $P$, and that element is different from $k$; while if $k_1,l_1$ are a disjoint, and $k_2,l_2$ are disjoint, then $k_1 \cup l_1; k_2 \cup l_2$ belong to the same element of $P$, and that element is different from $k$. We may say that $P$ preserves natural addition and subtraction. To extend this to make $P$ preserve natural multiplication, we need to make $P$ a partition on $\mathcal P (\mathcal P(\zeta)) \setminus \{\varnothing\}$, and additionally demand that if $l_1,l_2 \subset k$ and each is pairwise disjoint, then $\bigcup l_1, \bigcup l_2$ belong to the same element of $P$, and that element is different from $k$ if $l \neq 1$ and different from $l$ if $k \neq 1$, where $1$ is the set of singleton subsets of $\zeta$. For the latter case, we need each element $h$ of $P$ not just have infinitely many subsets of $\zeta$ as elements but also they must have infinitely many subsets of $\mathcal P(\zeta)$ that are not subsets of $\zeta$.
Is this consistent with $\sf ZFC$?
If yes, then can we have such partitions for every infinite cardinal?
