Isn't $\epsilon_{ij}$ an isotropic, rank-2 tensor?
I think @mikestone's comment is correct. The MathWorld site is talking about tensors in 3 dimensions, not 2D. In 3 dimensions (actually, in all dimensions $\geq3$), it can be shown that any isotropic rank-2 tensor is proportional to the identity ($\delta_{ij}$), see Richard Fitzpatrick's notes here, for example.
In two dimensions, there are two rank-2 isotropic tensors, $\delta_{ij}$, and what you have called $\epsilon_{ij}$. Here's a quick way to list all the isotropic tensors in 2D that I have drawn heavily from the fantastic analyses here and here. There are two equivalent ways to define an isotropic tensor: one is the way you have defined it, saying that $A$ is an isotropic tensor iff $$A = R\cdot A\cdot R^T,$$
but an equivalent way is to say that $A$ is isotropic iff it commutes with the generators of rotations, $L_i$, for example $[A,L_z] = 0$. Now, let's look at this condition in 2D. It turns out -- if you do the calculations -- that $$L_z = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},$$ precisely (the negative of) what you called $\epsilon_{ij}$!
Now consider an arbitrary 2D tensor $A$, and demand that it satisfy the commutation relation given above:
$$\Bigg[ \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}, \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\Bigg] = 0,$$
and you should be able to see that this just means such an arbitrary isotropic tensor in 2D should look like: $$A = \begin{pmatrix}a & -b \\ b & a\end{pmatrix},$$ and you should be able to see two things:
An arbitrary isotropic tensor in 2D is itself a rotation, since $$A = \frac{1}{\sqrt{a^2+b^2}}\begin{pmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix}, \quad \quad \alpha=\arctan{\left(\frac{b}{a}\right)}$$
An arbitrary isotropic tensor in 2D can be written as: $$A_{ij} = a \delta_{ij} -b \epsilon_{ij}.$$
Thus, there are precisely two independent isotropic rank-2 tensors in 2D, $\delta_{ij}$ and $\epsilon_{ij}$.
Note: I've not spoken too much about generators and how to derive $L_z$ in 2D since the answer would get too long, but I'd be happy to explain it if necessary.