Yes, for any closed Gaussian surface (including irregular surfaces), this law holds.
But we consider only symmetrical surfaces (sphere, cylinder, etc.) in academics as it is easy for performing the integration.
Let us consider the example which you have given:
Black dot represents a point charge.
Here, in order to find the net flux through the surface you must integrate each
E.dA, which is not so easy. The advantage with symmetry is that E is always constant, and dA is always perpendicular to the surface so that you may integrate easily. Even in this case, if you integrate E.dA, you will end up with the same result.
Gauss law even applies for irregular surfaces :
The only problem is to integrate. You would have the same result.
How to think without mathematics :
As we know, flux is the measure of number of lines of electric field passing through the Gaussian surface. Even if you move the charge anywhere within the surface, the number of field lines won't change, right?
All the Gauss laws(of electrostatics, magnetism, gravitation), field vectors, inverse square laws can be derived from the GAUSS LAW OF DIVERGENCE.
It's so exciting!!!
The electric field around the charges will have rotational symmetry about the line joining them, so the problem can be reduced to 2D and the task of proving that the locus is a circle.
The only physics here is getting $r_1=ar_2$. The rest is geometry. One proof relates to the Apollonian Circles Theorem.
Using co-ordinate geometry :
Suppose the charges are located at A(0,0) and B(d,0) where d=AB is the fixed distance between them. The distances AP, BP of some point P(x,y) from A,B are given by
$AP^2=x^2+y^2$
$AB^2=(d-x)^2+y^2$.
Set $a^2AP^2=BP^2$ to get an equation for the locus of points P. This has the form
$x^2+y^2-2fx-2gy+c=0$
which is the equation of a circle.
This problem is the inverse of finding the image charge A of a point charge B in a grounded conducting sphere. See Griffiths 2007, Problem 3.7.
I think Icchyamoy is wrong : the non-zero potential surfaces are not spherical. However, if a 3rd charge of suitable magnitude and position is introduced, the surface with any non-zero potential can be made spherical by a suitable choice of the magnitude and position of the 3rd charge. See A point charge near a conducting sphere.
Best Answer
You are correct that two equipotential surface are nothing as spaced out. At dx distance there will be another equipotential surface.
I think your textbook must be saying that the distance between potential surfaces differing by same potential is not constant.
Like the distance between surface of 2V and 4V will not be same as that between 4V and 6V.