The ZX calculus doesn't have a Y rotation nodes. All Y rotations must be decomposed into a series of X and Z rotations.
The reason for this asymmetry comes down to the fact that $Y \neq Y^T$ whereas $X=X^T$ and $Z=Z^T$. This is relevant because travelling around a Bell pair transposes gates. For all single qubit gates $U$, is is the case that $U_a(|0_a0_b\rangle + |1_a1_b\rangle) = U_b^T(|0_a0_b\rangle + |1_a1_b\rangle)$. This generalizes to multi-qubit gates. Note that this doesn't work if you use $U_b$ instead of $U_b^T$.
One of the very common things to do in a ZX graph is slide a node around a U turn. That's equivalent to moving a gate to the other side of a Bell pair. If Y nodes were allowed, this U turn would suddenly need to involve changing the node to be its transpose. You would lose the property that the layout of the graph was irrelevant to its function, or you would need a node that had an orientation which could be mirrored. When decomposing Y into X and Z rotations the need to transpose is naturally dealt with by the fact that the ordering of the decomposition gets flipped around as part of going around the U turn.