No.
There is no way (either with or without the Biot-Savart law!) to argue that $A=0$. In fact $A=0$ is a boundary condition, and it therefore does not have to be true.
Details.
The Maxwell equations can generally be applied within any big volume $V$ as long as you specifiy what $\vec E$ and $\vec B$ are on the boundary $\partial V$, and the Biot-Savart law is a result of the Maxwell equations.
If you imagine that your solenoid is oriented "up-down" and we put a North pole of a very big (much bigger than our room) round magnet somewhere underneath the floor, and a South pole of a very big round magnet somewhere above the ceiling, then (still talking in the case where V is the infinite 3D space!) we will essentially be adding a field $\vec B = B_0~\hat z$ to the field of the solenoid within the room, without messing with the existing rotational symmetry. Therefore all of your arguments about rotational symmetry still go through, but we find out that $A \ne 0$ in the room itself, rather $A = B_0$. Now if the room is much much larger than the solenoid therein, we see that the field is approximately $B_0~\hat z$ out on the boundary of this room and the Maxwell equations will give us this same solution if we remove the magnets but insist that we're only concerned with the field within this room, subject to this boundary condition that the field is $B_0~\hat z$ out on the boundary.
As we systematically allow increase the size of this room with those boundary conditions, we find a valid limit where the field "out at infinity" is $B_0~\hat z$. In fact the Maxwell equations allow us to fill space with any field which satisfies those equations in vacuum: so there are solutions where there is in fact a homogeneous plane wave of light permeating all of space, oscillating in the $x,y-$directions and travelling in the $z$-direction, too. The Maxwell equations don't say that this is invalid, because they cannot say that this is invalid. All they can say is that these do not have zero fields at the boundary.
However, if we do force the fields to be zero at infinity, the existence and uniqueness theorems tell us that there is just one solution of the Maxwell equations which doesn't have this pathological behavior way out there. Even better, that boundary condition is composable, so that we can take two point charges and their individual fields (assuming that said fields go to 0 at infinity) and construct the field for both charges together, by simply vector-summing the two fields. This is only possible because at the boundary we have that this reduces to $0 + 0 = 0$, so we now have a solution for both point charges when the field goes to 0 out at infinity.
And in fact this prescribes also the best way to approach these problems for large volumes $V$: get a solution which, with no charges inside $V$, enforces the (non-zero) boundary conditions; then superimpose all of the zero-boundary-condition results for all of your point charges and currents and whatnot on this background field to find the "final field" that you want.
Best Answer
Formula 1 is the contribution to the magnetic field from the small segment of the coil of length $dl$ shown in the lower portion of Figure 1. That piece of coil has current $dI = N I\, dl/l$ flowing through it where $N$ is the number of turns in the coil, $I$ is the current per turn, and $l$ is the length of the coil. From the hyperphysics site (or by integrating the Biot-Savart law), the axial field from this loop is: $$B_z = \frac{\mu_0}{4\pi}(2\pi r) dI\frac{ r}{(r^2+z^2)^{3/2}} = \frac{\mu_0 r^2 dI}{2 y^3}$$ where $y=(r^2+z^2)^{1/2}$ is the distance from the measurement point to the rim of the loop and $r$ is the coil radius. Substituting the formula for $dI$ gives you the first equation.
The second equation is the integral of the above over the length $l$. The integral is first converted to an integral over angle $\theta$ via $\sin \theta = r/y$ and $\sin\theta\,dl = y\,d\theta$: $$B = \int\frac{\mu_0 r^2 dI}{2 l y^3} = \frac{\mu_0 N I}{2 l} \int \frac{r^2 dl}{y^3} = \frac{\mu_0 N I}{2 l} \int_{\theta_1}^{\theta_2} \sin \theta \, d\theta$$