A contradiction between Biot-Savart and Ampère-Maxwell Laws?
The short answer is that the case of a finite wire violates one of the postulates of magnetostatics, which is that $\nabla \cdot \vec j = 0$. In cases when the current has sinks, the Biot-Savart law is not equivalent to the Ampere law, but to a "magnetostationary" Maxwell-Ampére law.
Hence, in this rather special case you get the same result from the Biot-Savart as well as the Maxwell-Ampére law. It is not by coincidence.
Sketch of a proof follows, for gory details see Griffiths' Introduction to electrodynamics. (Some details are on wikipedia)
The Biot-Savart law can be equivalently written as $$ \vec B(\vec{r}) = \frac{\mu}{4 \pi} \nabla \times \int \frac{\vec{j}(\vec{r}')}{|\vec{r}-\vec{r}'|} d^3 r' $$ It is important to remember that the curl acts only on the unprimed r. We can now take a curl of the above equation, use the curl curl formula, realize that the laplacian of $1/r$ is proportional to the delta function, and use a few other tricks to obtain $$\nabla \times \vec{B} =\frac{\mu}{4 \pi} \nabla( \int \frac{ \nabla' \cdot \vec{j}}{|\vec{r}-\vec{r}'|} d^3 r' ) + \mu \vec j$$ Where the primed nabla acts on the primed r. When the current is divergenceless, we then simply obtain the Ampere law in differential form.
However, if the current has a nonzero divergence and satisfies charge conservation we have $\nabla \cdot \vec{j}= -\partial_t \rho$. If we then assume that $\partial_t B=0$ ("magnetostationarity") the Gauss' law and other laws of electrostatics are unchanged. Then, we can use the electrostatic solution of the electric potential $\phi$ to easily derive that $$ \frac{\mu}{4 \pi} \nabla( \int \frac{-\partial_t \rho}{|\vec{r}-\vec{r}'|} d^3 r )=-\mu \epsilon \nabla( \partial_t \phi) $$ But we can also switch the order of derivatives and use $\vec E = -\nabla \phi$ to finally see that in the special "magnetostationary" case with conservation of charge the Biot-Savart law will be equivalent to the full Maxwell-Ampére law $$\nabla \times \vec{B} = \mu (\vec j + \epsilon \partial_t \vec E )$$
Note that the practical occurence of cases where the Maxwell contribution to Ampére law is non-negligible while the Faraday induction is negligible is very little. One should thus understand the "magnetostationary" validity of the Biot-Savart law as more of a curiosity.
For instance in your case the charges would have to be very large and the conductor between them a very bad conductor. As the two spheres become connected, an electromagnetic wave emerges. Only as the wave leaves the system and the stationary current is well established we can use the Biot-Savart law.
Yes, the result is correct provided the discharge is slow enough so that the electric field in the space near the point of interest and around the wire is accurately given by the Coulomb formula:
$$ \mathbf E_C(\mathbf x)=\frac{1}{4\pi \epsilon_0}\int \rho(\mathbf x') \frac{\mathbf x-\mathbf x' }{|\mathbf x-\mathbf x'|^3} d^3\mathbf x'. $$
This seems to be a very common situation - electric field near low-frequency AC electric circuits made of resistive and capacitive elements should be almost Coulombic(plausible, since the concept of potential is used successfully to analyze their operation$^1$).
In such situations, the Biot-Savart formula is applicable and gives accurate result. Here is the reason why.
The exact Maxwell-Ampere equation is
$$ \nabla \times \mathbf B = \mu_0 \mathbf j +\mu_0 \epsilon_0 \partial_t \mathbf E.~~~(*) $$
Let us denote the field given by the Biot-Savart formula as $$ \mathbf B_{BS}(\mathbf x) = \int \frac{\mu_0}{4\pi} \frac{\mathbf j(\mathbf x') \times (\mathbf x-\mathbf x')}{|\mathbf x-\mathbf x'|^3}\,\mathrm d^3\mathbf x'. $$
We will show that this field solves the exact equation (*) if electric field is given by the Coulomb formula.
Curl of this field can be expressed as
$$ \nabla \times \mathbf B_{BS}(\mathbf x) = \frac{\mu_0}{4\pi} \int \nabla_{\mathbf x} \!\times \big(\mathbf j(\mathbf x') \times \mathbf F (\mathbf x,\mathbf x')\big) d^3\mathbf x' $$ where $$ \mathbf F (\mathbf x,\mathbf x') = \frac{\mathbf x-\mathbf x' }{|\mathbf x-\mathbf x'|^3}. $$ The integral can be transformed into
$$ \int \mathbf j 4\pi \delta(\mathbf x'-\mathbf x) d^3\mathbf x' + \int (\nabla'\!\cdot \mathbf j) \mathbf F d^3\mathbf x' - \int\nabla' \cdot (\mathbf j \mathbf F) d^3\mathbf x'. $$
Let us choose the integration region in such a way that $\mathbf j = \mathbf 0$ everywhere on its boundary (moving charges are not crossing the boundary). Then the third integral is zero (this can be shown using the Gauss theorem). The divergence in the second integral can be expressed as $-\partial_t\rho$ based on the equation of continuity. Going back to curl of $\mathbf B_{BS}$, we obtain
$$ \nabla \times \mathbf B_{BS}(\mathbf x) = \mu_0 \mathbf j(\mathbf x) + \epsilon_0\mu_0\partial_t \mathbf E_C. $$
Now we can see that $\mathbf B_{BS}$ solves the exact Maxwell-Ampere equation (*) as long as the electric field is given by the Coulomb formula.
So, the Biot-Savart formula is not limited to static currents as people sometimes assume, but applies to changing currents as well. It is therefore more general than the integral Ampere formula.
$^1$ Except for phenomena of electromagnetic induction - near coils and moving magnets the electric field is rotational which cannot be described by the Coulomb formula.
Implicitly, what you are doing in this problem is taking the limit $\epsilon_0\rightarrow 0$ in an electrodynamics problem.
Let $q_+(t)$ and $q_-(t)$ be the two charges of the spheres at time $t$: $q_+(0)=Q$, $q_-(0)=-Q$. Since the velocity of light is (large but) finite, it will take some finite time for charge to move through the wire, so $|q_-(t)|-|q_+(t)|\geq 0$, and the difference is in the wire as a charge density $\rho(t,x)$. The current density in the wire is then $I(x)=v(x)\rho(x)$, where $v(x)$ is the velocity of the charges in the wire.
The version of Ampere-Maxwell that you have to use is w.r.t. $q_+, q_-,\rho$ and the electric fields and currents these two induce. Now, we take the limit $\epsilon_0\rightarrow 0$. What happens to the quantities in the problem? First off, at $\epsilon_0=0$, the Coulomb force is infinite, so $v(x)\rightarrow \infty$, and similarly, because the charge moves through the wire infinitely fast we get $|q_-(t)|-|q_+(t)|= 0$ and,therefore, $\rho(x)=0$. Now, the current density $I(x)$ should go to $dq/dt$ so that the continuity equation is satisfied. This explains that you use the correct current density.
The induced electric field is equal to the field by the charges $q_-,q_+$, that you use, plus radiative effects. These radiative effects (See the Lienard-Wiechert potentials for an example) vanish in the limit $\epsilon_0\rightarrow 0$, which takes $c\rightarrow \infty$. So for the electric field you only have to use the static component, like you did. This static component is infinite, but the infinity cancels: the $\epsilon_0/\epsilon_0$ that you crossed out takes care of that.
For full mathematical rigor, you have to be a bit careful about how you take all these limits, but I promise you it checks out.
Equivalently, if you look up the the Lienard-Wiechert potentials on wikipedia, you find the relativistic generalisation to the Biot-Savart law. If you take $\epsilon_0\rightarrow 0$ in this expression, you recover the Biot-Savart law. This proves, in an admittedly roundabout manner, that both approaches are equivalent.
If you read between lines of the above argument, what is hidden there is a proof that the Biot-Savart law only neglects the radiating fields. Variations in the static Coulomb part of the electric field are correctly taken into account by Biot-Savart, which is why the two methods agree.