How to find the magnetic field outside an infinite solenoid without Biot-Savart law? Let's consider an infinite solenoid, with $n$ loops per unit length, radius $a$, which carries a current $i$. Using symmetries and Ampère's law, one can find that the magnetic field is ortho-radial, and that
$$\begin{cases}
B_{int} = \mu_0ni + A \\
B_{ext} = A
\end{cases}$$
where $A$ is a constant that we must now determine.
Using Biot-Savart law, I find that $A=0$, however I'd like to do it without this law. Is there a way to do so ?
 A: 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.
A: We typically require that the field approaches zero at an infinite distance from the solenoid. This leads immediately to $A=0$.
A: Consider a rectangular amperian loop abcd. Along cd the field is zero. Along transverse sections bc and ad, the field component is zero. Thus these two sections make no contribution. Let the field along ab be B. Thus,the relevant length of the amperian loopis, L=H.
let n be number of turns per unit length then the total number of turns is nh. The enclosed current is I(nh), where I is the current in the solenoid. From ampere circuital law 
BL=uI
Bh=uI(nh)
B=unI
(Here u is free space permeability)
A: Much of what I write can be read in greater detail in Griffiths.
Let solenoid be along z axis. B can only have a z component throughout space. By symmetry you can kill radial component and by using ampere's law you can kill the tangential component. 

Look at loop 1. Ampere's law applied here easily gives $B(a)=B(b)$.
Now, its easy to say why $B(a)= B(b) = 0$. Let b tend to infinity. 
If field at infinity is finite, it means, for every finite length  of solenoid, there is infinite energy stored in magnetic field,
 Given by $$\frac{1}{2\mu_0}\int B^2\; dV$$
Comparing energy in whole space due to infinitely long solenoid with this energy is like comparing a plane to a 3D space. They are totally different. 
