It is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems.
Correct, they are equivalent (assume no GR, and no QM) in the sense that if the integral versions hold for any surface/loop then the differential versions hold for any point, and if the differential versions hold for every point then the integral versions hold for any surface/loop. (This also assumes you write the integral versions in the complete and correct form with the flux of the time partials of the fields and/or with stationary loops.)
Suppose there is a time-varying current in a wire $I(t)$ and I wish to find the fields a long way from the wire.
Ampere's Law is correct, but it is not also helpful. If you know the circulation of $\vec B$, you can use it to find the total current (charge and displacement). If you know the total current (charge and displacement), then you can find the circulation of $\vec B$. But solving for $\vec B$ itself is hard unless you have symmetry.
What about using Ampere's law in integral form? What is the limit of its validity?
It is completely valid, but it might not be helpful. When you write:
$$ \oint \vec{B}(r,t)\cdot d\vec{l} = \mu_0 I(t) + \mu_0 \iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{a}$$
then the $t$ that is used on each side of the equation is exactly the same.
When $I(t)$ changes, then $\vec B$ field nearby changes quickly, and when there is a changing $\vec B$ field there is a circulating electric field, so as the region of changing $\vec B$ field expands, so does the region of newly circulating electric fields. Both expand together. Eventually the expanding sphere of changing $\vec B$ field and changing circulating electric fields finally starts to reach the Amperian loop (together), and only then does the circulation of $\vec B$ on the far away Amperian loop change. If there was just one change in $I(t)$, then the expanding shell of changing electric fields continues expanding, and you are stuck with the new value of the circulating $\vec B$ field, based on the current that changed a while back.
So, to solve for $\vec B$, you'd need both $I$ and $\partial \vec E /\partial t$ and the latter you need for all the empty space on a surface through the Amperian Loop. Maxwell's equations don't have limited validity and do not need to be modified. They just aren't always as useful as you might want them to be.
I think your limiting procedure should work.
For the homogeneous Maxwell equations, when we restrict to a surface of constant $x$, the pull back of the field strength looks something like
$$\iota^* F = E^\perp_i dt \wedge dx^i + B_x dy \wedge dz,$$
where $E^\perp$ is the component of $E$ in the $yz$-plane. Now your timelike pillbox integral will have three contributions, one from the sides of the cylinder, and two from the top and bottom. The tangent vectors of the sides are $\partial_t$ and $v^\mu$, the tangent vector of the curve bounding the surface in the $yz$-plane. So the side contribution gives
$$\int dt\oint \mathbf{E}\cdot\mathbf{v}ds,$$
where $s$ is a parameter on the curve. The top and bottom of the integral comes entirely from the $B_x dy \wedge dz$ term, and is just the usual flux of $B$ through the surface. Since the total contribution is zero, we have
$$\int_t^{t+h} dt \oint \mathbf{E}\cdot\mathbf{v}ds + \int_{t+h} \mathbf B\cdot d\mathbf{A} - \int_{t} \mathbf B\cdot d\mathbf{A} = 0$$
Now if you want to take the limit as $h\rightarrow 0$, you can Taylor expand each term and look at the first order in $h$ terms. The first integral just becomes the line integral by the Fundamental Theorem of Calculus. The $\mathbf{B}$ flux integrals can also be seen straightforwardly to approach the time derivative of the flux at first order in $h$. Thus the result is
$$\oint \mathbf{E}\cdot \mathbf{v} ds + \frac{\partial}{\partial t}\int_t \mathbf{B}\cdot d\mathbf{A} = 0$$
Applying basically the same procedure to the dual field strength gives
$$\oint \mathbf{B}\cdot \mathbf{v} ds -\frac{\partial}{\partial t}\int_t \mathbf{E}\cdot d \mathbf{A} = \int_t \mathbf{j}\cdot d\mathbf{A}.$$
Best Answer
The hand-wavy way to do it is to consider a wave solution like the one below, and apply Faraday's law to loop 1, and Ampere's law to loop 2:
If you make the loops narrow enough, i.e., their widths are $dx$, then $$\oint_1\!\vec{E}\cdot \vec{ds} = -\frac{d\Phi_B}{dt} \to \frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{dt}$$ $$\oint_2\!\vec{B}\cdot \vec{ds} = \varepsilon_0\mu_0\frac{d\Phi_E}{dt} \to \frac{\partial B_z}{\partial x} = -\varepsilon_0\mu_0\frac{\partial E_x}{dt}$$ Now differentiate the first equation wrt $t$ and the second wrt $x$, and combine to obtain the wave equation: $$\varepsilon_0 \mu_0 \frac{\partial^2 E_x}{\partial t^2} = \frac{\partial^2 E_x}{\partial x^2}$$