Isotropic can be rephrased as "invariant" to the direction. Its value should not depends on the angle under which you see the image.
In other words: if you have a regular function $g(x,y)$ (a continuous 2D image) expressed in the standard $(X,Y)$ referential, or in a $\theta$-rotated referential $(X \cos \theta + Y \sin \theta,-X \sin \theta + Y \cos \theta)$, their Laplacians should be equal (at least close enough). The Laplace operator (in its continuous expression) is rotationally invariant (and more generally, invariant under orthogonal transformations). This is a classical result of rotational invariance for operators, and answered in SE.maths Show Laplace operator is rotationally invariant:
$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}$$
where
$$u = x \cos \theta + y \sin \theta$$
$$v = -x \sin \theta + y \cos \theta$$
Now, discrete images are sampled along a cartesian grid that is not really compatible with arbitrary rotations. There are many discrete approximation candidates. Let us focus on the $3\times 3$ kernels. The two most standard ways of linking pixels for 2D images is the 4-connectivity (or Von Neumann neighborhood: north, east, south, west) and the 8-connectivity (or Moore neighborhood, adding corners NE, NW, SE, SW). From these connectivities you get the two most classical discrete Laplacians:
$$\begin{bmatrix}
0 & -1 & 0\\-1 & 4 & -1\\0 & -1 & 0
\end{bmatrix}$$
and
$$\begin{bmatrix}
-1 & -1 & -1\\-1 & 8 & -1\\-1 & -1 & -1
\end{bmatrix}$$
The first one is invariant to $90°$ rotations, the second one is less sensitive to $45°$ angles. Yet, many people are tried to derive better versions, see form instance Constructing an "isotropic" Laplacian operator. Back to the works of A. Rosenfeld, in his paper Optimally isotropic Laplacian operator, 1999:
Laplacian operators used in the literature for digital image
processing are not rotationally invariant. We examine the anisotropy
of $3\times 3$ Laplacian operators for images quantized in square
pixels, and find the operator which has the minimum overall
anisotropy.
he finds an optimum:
$$-\frac{1}{4}\begin{bmatrix}
1 & 2 & 1\\2 & -12 & 2\\1 & 2 & 1
\end{bmatrix}$$
"by minimizing
the anisotropy over all angles and all edge distances". And there are many other works with larger kernels, under different penalties.
Additional information: Role of the rank of the filter mask matrix in image processing?