The Biot-Savart law says that under magnetostatic conditions ($\frac{\partial}{\partial t}\rightarrow 0$),
$$\mathbf B(\mathbf r) = \frac{\mu_0}{4\pi}\int \frac{\mathbf J(\mathbf r') \times (\mathbf r - \mathbf r')}{|\mathbf r - \mathbf r'|^3} dV'$$
Noting that
$$ \frac{\mathbf r - \mathbf r'}{|\mathbf r - \mathbf r'|^3} = \nabla \left(\frac{1}{|\mathbf r - \mathbf r'|}\right)$$
where $\nabla$ refers to differentiation by the unprimed coordinates, this can be written
$$\mathbf B(\mathbf r) = \nabla \times\frac{\mu_0}{4\pi} \int \frac{\mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}dV'$$
Taking the curl of this and using the fact that $\nabla \times (\nabla \times \mathbf F) = \nabla(\nabla \cdot \mathbf F) - \nabla^2 \mathbf F$,
$$\nabla \times \mathbf B(\mathbf r) = \nabla\left(\int J(\mathbf r')\cdot \nabla\left[\frac{1}{|\mathbf r - \mathbf r'|}\right]dV'\right) - \nabla^2 \frac{\mu_0}{4\pi} \int \frac{\mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}dV'$$
Noting that
$$\nabla\left[\frac{1}{|\mathbf r - \mathbf r'|}\right] = -\nabla'\left[\frac{1}{|\mathbf r - \mathbf r'|}\right]$$
we can integrate the first term by parts to obtain
$$\nabla\left(\int \nabla' \cdot \left[\frac{\mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}\right]dV' - \int\frac{\nabla' \cdot \mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|} dV'\right)$$
The first term is a surface term, and vanishes if we assume that $\mathbf J(\mathbf r') \rightarrow 0$ as $|\mathbf r'| \rightarrow \infty$. The second term vanishes because according to the continuity equation, $\nabla\cdot\mathbf J = -\frac{\partial \rho}{\partial t} = 0$ in magnetostatics. This leaves us with
$$\nabla \times \mathbf B(\mathbf r) = \nabla^2 \frac{\mu_0}{4\pi} \int \frac{\mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}dV'$$
and since
$$\nabla^2 \frac{1}{|\mathbf r - \mathbf r'|} = 4\pi \delta^{(3)}(\mathbf r - \mathbf r')$$
we have
$$\nabla \times \mathbf B = \mu_0 \mathbf J$$.
Again, Biot-Savart is valid only under magnetostatic conditions, and therefore so is this version of Ampere's law. It would be nice to relax these conditions and re-do this derivation more generally, but we don't yet know what to replace Biot-Savart with.
Instead, let's see how this version of Ampere's law fails when we go to general electrodynamics. Clearly since $\mathbf J \propto \nabla \times \mathbf B$, we have that $\nabla \cdot \mathbf J = 0$. However, according to the general continuity equation, $\nabla \cdot \mathbf J = -\frac{\partial \rho}{\partial t}$.
To fix this, let's assume that we need a new term, so
$$\nabla \times \mathbf B = \mu_0 \mathbf J + \mathbf G$$
for some vector field $\mathbf G$. Taking the divergence of both sides yields
$$ 0 = - \mu_0 \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf G$$
$$\implies \nabla \cdot \mathbf G = \mu_0 \frac{\partial \rho}{\partial t}$$
From Gauss' law for electric fields, we know that $\rho = \epsilon_0 \nabla \cdot \mathbf E$, and so
$$\nabla \cdot \mathbf G = \epsilon_0 \mu_0 \frac{\partial}{\partial t} \nabla \cdot \mathbf E = \nabla \cdot \left(\epsilon_0 \mu_0 \frac{\partial}{\partial t}\mathbf E\right)$$
and so we can simply postulate that
$$\mathbf G = \epsilon_0\mu_0 \frac{\partial}{\partial t} \mathbf E$$
so
$$\nabla \times \mathbf B = \mu_0 \mathbf J + \epsilon_0\mu_0 \frac{\partial}{\partial t} \mathbf E$$
This was Maxwell's correction to Ampere's law, and it has been validated over and over by experiment.
In summary, magnetostatics + Biot-Savart gives us $\nabla \times \mathbf B = \mu_0 \mathbf J$. Predictably, this fails when we leave the domain of magnetostatics, and in particular is inconsistent with the continuity equation. We don't know how to generalize Biot-Savart, but patching up the problem with the continuity equation in the simplest possible way yields the correct Ampere's law, $\nabla \times \mathbf B = \mu_0 \mathbf J + \epsilon_0 \mu_0 \frac{\partial}{\partial t}\mathbf E$.
From this, we can work backward to find the correct generalization of Biot-Savart; this is one of Jefimenko's equations.
EDIT:
Returning to the original derivation after eliminating the surface term (but before sending $\nabla'\cdot \mathbf J(\mathbf r')\rightarrow 0$), we have
$$\nabla \times \mathbf B(\mathbf r) = \mu_0 \mathbf J(\mathbf r) - \frac{\mu_0}{4\pi}\int \frac{\nabla '\cdot \mathbf J(\mathbf r')}{|\mathbf r - \mathbf r'|}$$
Under the conditions for which Biot-Savart holds, the latter term is equal to zero; however, we can be bold and throw those restrictions aside just to see what happens. Under general conditions, $\nabla \cdot \mathbf J = -\frac{\partial \rho}{\partial t}$, so that term becomes
$$\frac{\mu_0}{4\pi} \nabla \int \frac{\partial \rho}{\partial t} \frac{1}{|\mathbf r - \mathbf r'|} dV'= \frac{\partial}{\partial t} \frac{\mu_0}{4\pi} \nabla \int\frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|}$$
Defining $$ \phi(\mathbf r) = \int \frac{\rho(\mathbf r')}{4\pi \epsilon_0 |\mathbf r-\mathbf r'|}$$
and letting $\mathbf E = -\nabla\phi$, this becomes
$$\nabla \times \mathbf B = \mu_0 \mathbf J + \epsilon_0 \mu_0 \frac{\partial}{\partial t} \mathbf E$$
What we did here - simply disregarding the conditions under which Biot-Savart is applicable and plugging in the more general continuity equation - is morally the same as Maxwell's addition of the extra term to compensate for the nonzero divergence of $\mathbf J$.
Note also that we have glossed over how to go from magnetostatics to electrodynamics $\big(\mathbf J(\mathbf r) \rightarrow \mathbf J(\mathbf r,t), \rho(\mathbf r)\rightarrow \mathbf \rho(\mathbf r,t)\big)$. Simply plugging in a $t$ to Biot-Savart and letting it "go along for the ride" is insufficient; working backwards from the full Maxwell's equations demonstrates the need to introduce the retarded time $t_r = t - \frac{|\mathbf r - \mathbf r'|}{c}$, indicating that Biot-Savart is genuinely wrong for electrodynamics.
