Let me please utilize a different notation and use the symbol $\mathbf{\hat{\theta}}$ for the unit vector around the wrapped sample (instead of $\mathbf{\hat{x}}$, since it is along an angular coordinate after wrapping). So the gauge transformation is:
$$\mathbf{A} \rightarrow \mathbf{A} + A_0 \mathbf{\hat{\theta}}$$
Using the Cartesian coordinates in the plane of the loop $x,y$ (Please notice again that $x$ is not the same $x$ used by Laughlin), then by simple trigonometry, we have:
$$\mathbf{\hat{\theta}} = \frac{-y\mathbf{\hat{x}} + x\mathbf{\hat{y}} }{x^2+y^2}$$
This vector function is indeed a gauge transformation, since
$$A_0\mathbf{\hat{\theta}} = A_0\frac {-y\mathbf{\hat{x}} + x\mathbf{\hat{y}} }{x^2+y^2 } = A_0 \mathbf{\nabla} \tan^{-1}(\frac{y}{x})$$
The above relation is valid everywhere except at the origin, where right hand side is ill-defined.
It is easy also to check that the magnetic field "due" to this gauge transformation is vanishing except at the origin:
$$\mathbf{B} = A_0\mathbf{\nabla} \times \mathbf{\hat{\theta}} = 0 $$
However, this magnetic field has a non-zero flux which can be easily calculated:
$$\Phi = \int d\mathbf{S}.\mathbf{B} = A_0\oint \mathbf{\hat{\theta}} . d\mathbf{\hat{r}}$$
where Stokes theorem was used and $\mathbf{\hat{r}} = dx \mathbf{\hat{x}}+ dy \mathbf{\hat{y}}$ is the line element along the integration trajectory. Writing the line integral in Cartesian coordinates, we obtain:
$$\Phi = \int d\mathbf{S}.\mathbf{B} = A_0\oint \frac {-ydx+ xdy }{x^2+y^2 } $$
The integral can be easily solved in polar coordinates: $x = r \cos \theta, $ $y= r \sin \theta$:
We have:
$$\Phi =A_0\int_{0}^{2 \pi} d\theta = 2 \pi A_0$$
(In Laughlin article, the circumference of the loop is denoted by $L_x$). Please observe that the integral does not depend on the radius of the loop. The source of the flux is a delta function singularity of the magnetic field in the center of the circle.
Thus it was established that this "gauge transformation" is a result of incorporation of a flux along the axis of the loop. This transformation, therefore describes a physically different situation. Actually, it is not a gauge transformation, since a gauge transformation should describe a physically identical situation. It is called a Large gauge transformation. It was discussed here many time on physics stack exchange, please see for example this question . This transformation is actually a symmetry and it is characterized by the property that its gauge function $\phi (\theta) = A_0 \mathbf{\hat{\theta}} $ is not a function on the circle, since $\phi(2\pi)-\phi(0) \neq 2 \pi n$ (except when $A_0$ is not an integer).
Please note also that the gauge potential in this question is an Aharonov-Bohm gauge potential, since its magnetic field vanishes except at singular points (A gauge potential satisfying this condition is called flat).
Update
Remark(1): I think that Laughlin wanted to keep his explanation to as simple as possible so he did not discuss the issue of the dependence of the phase on the trajectory. His main aim was to show that the gauge transformation becomes something different after wrapping the sample.
In the answer above, I likewise did not write all the factors in order to keep things simple. The interaction term in the action is $\int \frac{q}{\hbar c} \mathbf{A} \cdot d\mathbf{r}$. Now, the solution including all the factors (now written in polar coordinates) has the form: $\mathbf{A} = \frac{\Phi}{2 \pi r} \mathbf{\hat{\theta}}$. Please notice that $\mathbf{A}$ has units of [Flux/length] as it should. The line integral becomes: $\int {q}{\hbar c} \frac{\Phi}{2 \pi r} r d\theta= \frac{q \Phi}{\hbar c} $. As can be seen, the dependence on the radius cancels. The phase does not depend on the shape of the trajectory either.
The important thing to remember is that Berry phases have the dimensions of an angle, they cannot be proportional to quantities of dimensions of length. But there are two cases: If the phase depends on the integration trajectory (and therefore on the solid angle enclosed by the trajectory), it is called a geometric phase. The second case is when the Berry phase does not depend on the trajectory (as long as it encloses the origin). The latter is called a topological phase and this is the type of phase in the question.
Remark(2): Indeed, large gauge transformations occur , in general, when the manifold under consideration is compact. This is the case described by Laughlin, when the particle's trajectory became a circle instead of a straight line. They encode topological invariants of the manifolds under consideration. In our case it is the cohomology class of the gauge transformation.
I just want to draw your attention that this example by Laughlin is only the introduction to what is known now by the (very famous) name: "Laughlin's argument", where he uses this geometry described in the question to prove the quantization of the Hall conductance.
