A specific spherical harmonic is not rotationally invariant, and the Wikipedia article does not claim this. What is true is that the space of all spherical harmonics of a fixed degree is a finite-dimensional irreducible representation of the rotation group $\text{SO}(3)$, and that the spherical harmonics form particularly nice bases of these representations. (In particular, rotating a spherical harmonic gets you a linear combination of spherical harmonics.)
I'm not exactly sure what you are asking, however the following may be useful as a less verbose way of obtaining the same result.
The key fact is that the cross product of $A,B$ is the unique element $A\times B$ such that
$\langle x, A\times B \rangle = \det \begin{bmatrix} A & B & x\end{bmatrix}$, $\forall x$.
Let $Q$ be a rotation (ie, $Q^TQ = I$), then using the properties of $\det$ we have
$$\det \begin{bmatrix} A & B & x\end{bmatrix} =\det Q^T Q \det \begin{bmatrix} A & B & x\end{bmatrix} = \det Q \det \begin{bmatrix} QA & QB & Qx\end{bmatrix},$$
from which we obtain $\langle x, A\times B \rangle = \det Q \langle Qx, QA\times QB \rangle = \langle x, (\det Q) Q^T(QA\times QB) \rangle$.
Since this is true for all $x$, we have $A\times B = (\det Q) Q^T(QA\times QB)$,
or
$$Q(A\times B) = (\det Q) QA\times QB.$$
Thus, if $Q$ is a proper rotation ($\det Q = +1$), you have $Q(A\times B) = QA\times QB$ (ie, the cross product is invariant under proper rotations).
Your question posits that both $A,B$ are invariant under rotation (which seems like a fairly restrictive condition), in which case $A = QA$, $B= QB$, from which it follows that $Q(A\times B) = A\times B$, ie, $A\times B$ is invariant under rotation too (assuming a proper rotation, of course).
Best Answer
This is most easily proved without coordinates. The cross product is the unique vector that is orthogonal to both factors, has length given by the area of the parallelogram they form and forms a right-handed triple with them. These properties are all invariant under rotations, and thus so is the cross product.