Mathematically speaking, it is true that Maxwell's equations by themselves aren't the whole story; they're a set of PDEs for which one needs to specify boundary conditions separately if one wants to solve them. Boundary conditions can be well-motivated from a physical perspective in a given scenario, but they do not follow from the equations themselves.
As for Poisson's equation versus Coulomb's Law, no requirement of spherical symmetry is necessary. Start with Poisson's equation, and set the charge density to be that of a point charge, namely
\begin{align}
\rho(\mathbf x) = q\delta(\mathbf x - \mathbf x_0).
\end{align}
Next, use the vanishing curl of the electric field (which works in the absence of explicitly time-varying magnetic fields) to write $\mathbf E = -\nabla \Phi$ and plug this into Poisson's equation to obtain
\begin{align}
\nabla^2\Phi = -\frac{q}{\epsilon_0}\delta(\mathbf x-\mathbf x_0)
\end{align}
In other words, we want to determine $\Phi$ that is a Green's function for the Laplace equation. The general solution is
\begin{align}
\Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|} + F(\mathbf x)
\end{align}
where $F$ is a harmonic function, namely one which satisfies Laplace's equation. If we then invoke the boundary condition that the potential vanishes at infinity, then this forces $F$ to be identically zero, and we obtain
\begin{align}
\Phi = \frac{1}{4\pi\epsilon_0}\frac{q}{|\mathbf x - \mathbf x_0|}
\end{align}
which, upon taking the gradient, yields he electric field of a point charge which is essentially the content of Coulomb's Law.
As far as I can remember, the formula you obtain is right. You can make this "problematic" integral disappear by using the following identity, that we will call "curl theorem" :
$$\int\vec{\nabla}\times\vec{w}dV = -\int\vec{w}\times d\vec{S}$$
To show this is true, we are going to use the divergence or Green-Ostrogradski theorem, namely
$$\int\vec{\nabla}\cdot \vec{v}dV = \int \vec{v}\cdot d\vec{S}$$
Since the divergence theorem is a scalar identity while the curl theorem is a vector identity, we are going to need three distinct vector fields that we are going to denote $\vec{v}_i$. Now, we would want $\vec{\nabla}\cdot\vec{v}_i = (\vec{\nabla}\times\vec{w})_i$ to deduce an identity on the curl. Writing that in tensor notation :
$$\partial^k(v_i)_k=\epsilon_{ikl}\partial^k w^l$$
As we can see, it is sufficient to take $(\vec{v}_i)_k = \epsilon_{ikl}w^l$ and the relation will be satisfied. So, for such a vector field we have $\vec{\nabla}\cdot\vec{v}_i = (\vec{\nabla}\times\vec{w})_i$.
Applying the divergence theorem to $\vec{v}_i$ :
$$\int(\vec{\nabla}\times\vec{w})_idV = \int\vec{\nabla}\cdot\vec{v}_idV = \int\vec{v_i}\cdot d\vec{S} = \int (v_i)_k(d\vec{S})^k = \int\epsilon_{ikl}w^l(d\vec{S})^k = -\int(\vec{w}\times d\vec{S})_i$$
Thus giving a proof of the "curl theorem". Using it on your problematic integral :
$$-\frac{\mu_0}{4\pi}\int\nabla'\times\left(\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\right)\,d^3\mathbf{x}' = -\frac{\mu_0}{4\pi}\int\left(\frac{\mathbf{J}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\right)\,\times d\vec{S}'$$
Now, the volume integral is done on all of space, and provided you suppose that $\lim_{x'\rightarrow\infty}\frac{\vec{J}(x')}{|x-x'|} = 0$, it gives a 0 contribution. Why does this not add any crazy assumptions ?
For this limit to be non-zero, we must necessarily have that $|J(x)|$ tend to infinity. Indeed, suppose $J(x)$ is finite. Then, there is a constant $C$ such that $|J(x)|<C$. Then, $lim_{x'\rightarrow\infty}\frac{|J(x')|}{|x-x'|}<\lim_{x'\rightarrow\infty}\frac{C}{|x-x'|} = 0$. Thus, if we were to have this "extra" integral not vanish, we would be required to have an infinite current density at infinity, which seems to be not so physical.
Of course, all my derivation where done in the context of well-behaved functions. It won't work say for an infinitely small wire, as the current density becomes a distribution (using the dirac delta $\delta(x)$). I am not qualified enough to tackle this case rigorously, but I hope the explanation above gives an idea to why setting this integral to 0 is sensible.
Best Answer
This isn't really a "boundary conditon" problem in the sense that we aren't trying to take knowledge of $\vec{A}$ on some surface and extend it to a solution of Laplaces equation in some region, which has the surface as a boundary. The reason you can't do this is that you have no apriori reason to know how $\vec{A}$ should look for a given $\vec{J}$.
Rather you're trying to turn information about the sources, $\vec{J}$, into information about the field, $\vec{A}$. This requires actually inverting the $\nabla^2$ operator, which involves the use of Green's functions. For instance you could use the formula,
$$ A_x(\vec{x}) = \frac{\mu}{4\pi} \int \frac{J_x(\vec{x}') d^3\vec{x}'}{\left| \vec{x}-\vec{x}' \right|} .$$