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.)
Here is a longer derivation that requires mostly basic calculus.
The Laplacian is invariant under rotations. Specifically, if $f:\mathbb R^n\rightarrow \mathbb R^n$ and $\nabla^2 f(x)\;=\alpha$, then $\nabla^2 g\;
(R^{-1} x)=\alpha$ where $R$ is any rotation matrix ($R^TR=I$ and $\det(R)=1$) and $g(x) := f(Rx)$.
Now suppose that $f:\mathbb R^n\rightarrow \mathbb R^n$ has the property that $f(x)= h(||x||)$ where $h:\mathbb [0,\infty) \rightarrow \mathbb R$, $h$ is twice differentiable, and $||x||=\sqrt{\sum_i x_i^2}.$
At $x$, we can form an orthonormal basis $\{x/||x||, v_2, v_3, \ldots v_{n-1}\}$. Now
$$
\nabla^2 f(x) = \sum_{i=1}^n \frac{\partial^2 f}{\partial x_i^2}(x)
$$
and by invariance under rotation, we can use the basis $\{x/||x||, v_2, v_3, \ldots v_{n-1}\}$ to get
\begin{equation}
(1)\quad\quad\nabla^2 f(x)= h''(||x||) + \sum_{i=2}^n p_i
\end{equation}
where $p_i:= k_i''(0)$ and $k_i(\epsilon) := f(x+\epsilon v_i)$.
The fact that $f$ only depends on the radius implies that all of the $k_i$ are the same and for $i=2, 3, \ldots n$,
$$f(x+\epsilon v_i) = k_i(\epsilon) = k(\epsilon) = f(x+ \epsilon v_2)=h(\sqrt{||x||^2+\epsilon^2}).$$
Let $r=||x||$. Using calculus,
$$k'(\epsilon) = h'(\sqrt{r^2+\epsilon^2})\frac{\epsilon}{\sqrt{r^2+\epsilon^2}}, \quad \mathrm{and} $$
$$k''(\epsilon) = h''(\sqrt{r^2+\epsilon^2})\frac{\epsilon^2}{r^2+\epsilon^2} + h'(\sqrt{r^2+\epsilon^2})\frac{\sqrt{r^2+\epsilon^2} -\epsilon \frac{\epsilon}{\sqrt{r^2+\epsilon^2}} }{r^2+\epsilon^2}.$$
Thus $p_i:= k''(0) = h''(r)\cdot 0 + h'(r)\frac{r}{r^2}=h'(r)/r$. Applying this to equation (1) gives
$$
\nabla^2f(x)= h''(||x||) + \sum_{i=2}^n p_i = h''(r) + (n-1)\frac{h'(r)}{r}.
$$
Best Answer
In this 2-dimensional case, everything is much simpler, I agree. In fact you can even write down explicitly what a general rotation looks like. So, suppose you have two sets of coordinates; $(x,y)$ and $(u,v)$, where one is obtained from another by a rotation, say of angle $\phi$: \begin{align} \begin{cases} u &= x\cos \phi - y \sin \phi \\ v &= x \sin \phi + y \cos \phi \end{cases} \end{align} Now, using the chain rule, we find that \begin{align} \dfrac{\partial}{\partial x} &= \dfrac{\partial u}{\partial x} \dfrac{\partial }{\partial u} + \dfrac{\partial v}{\partial x} \dfrac{\partial}{\partial v} \\ &= \cos \phi \dfrac{\partial}{\partial u} + \sin \phi \dfrac{\partial}{\partial v} \end{align} and similarly, \begin{align} \dfrac{\partial}{\partial y} &= -\sin \phi \dfrac{\partial}{\partial u} + \cos \phi \dfrac{\partial}{\partial v} \end{align} Now, try to calculate $\dfrac{\partial^2}{\partial x^2}$ and $\dfrac{\partial^2}{\partial y^2}$ similarly, and then add them up. You should find in a few lines of algebra (after using $\sin^2 + \cos ^2 = 1$ a couple of times) that \begin{align} \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} = \dfrac{\partial^2}{\partial u^2} + \dfrac{\partial^2}{\partial v^2} \end{align}
Edit: Answering Question in comments
We have \begin{align} \dfrac{\partial^2 f}{\partial x^2} &=\dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial x} \right) \end{align} Now, temporarily define $g$ as \begin{align} g:= \dfrac{\partial f}{\partial x} = \dfrac{\partial u}{\partial x} \dfrac{\partial f }{\partial u} + \dfrac{\partial v}{\partial x} \dfrac{\partial f}{\partial v} = \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v} \end{align} So, \begin{align} \dfrac{\partial ^2 f}{\partial x^2} &= \dfrac{\partial g}{\partial x} \\ &= \dfrac{\partial u}{\partial x} \cdot \dfrac{\partial g}{\partial u} + \dfrac{\partial v}{\partial x} \cdot \dfrac{\partial g}{\partial v} \\ &= \cos \phi \dfrac{\partial g}{\partial u} + \sin \phi \dfrac{\partial g}{\partial v} \\ &= \cos \phi \dfrac{\partial }{\partial u} \left( \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v}\right) + \sin \phi \dfrac{\partial }{\partial v} \left( \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v} \right) \\ &= \cos^2 \phi \dfrac{\partial ^2 f}{\partial u^2} + 2\cos \phi \sin \phi \dfrac{\partial ^2 f}{\partial u \partial v} + \sin^2 \phi \dfrac{\partial ^2 f}{\partial v^2} \end{align} where in the last line, I expanded everything, and used equality of mixed partials. If you do a similar thing with $y$, you'll get a $-2 \sin \phi \cos \phi$ term instead.