In 3-space, one can interpret the 4 Maxwell equation as determining the relationship between the fields (the electric field vector and the magnetic field bivector) and all four types of possible sources.
But this is rather illusory. In relativity, the equations look quite different:
$$\begin{align*} \nabla \cdot F &= -\mu_0 J \\ \nabla \wedge F &= 0\end{align*}$$
where $F$ is the electromagnetic field bivector. The vector derivative $\nabla$ can only increase or decrease the grade of an object by 1. Since $F$ is grade 2, the divergence equation describes its relationship with a grade 1 source term (the vector four-current $J$). The curl equation describes its relationship to a grade 3 (trivector) source term (of which there is none).
The reason the 4 Maxwell equations in 3-space come out the way we do is that we ignore the timelike basis vector, which would unify the scalar charge density with the 3-current as the four-current, as well as unify the E field with the B field as a bivector. The relativistic formulation, however, is considerably more sensible, as it correctly presents the EM field as one object of a single grade (a bivector), which can only have two sources (vector or trivector). It just so happens that the EM field has no trivector source.
What if there were trivector sources? Well, as you observe, there would be magnetic charge density (monopoles), but there would also be quite a bit more. There would have to be magnetic current as well, which would add an extra term to the $\nabla \times E$ equation to fully symmetrize things.
If the differential form is fundamental, we won't get any current, but the integral form is fundamental we will get a current.
I'm not sure how you came to that conclusion, but it's not true. Both the differential and integral forms of Maxwell's equations are saying exactly the same thing. Either can be derived from the other, and both of them predict the exact same physical consequences in any situation.
Most physicists would say the differential form is more fundamental, but that's just an artifact of how we think about modern physics, in terms of fields which interact at specific points. It's really a philosophical issue, not a physical one, because it doesn't matter for the purpose of doing calculations which form you consider to be more fundamental.
In the specific situation you're asking about, with the solenoid, you will get a current in the loop around the solenoid. It may be easier to see that by using the integral form of Faraday's law, but the differential form makes the exact same prediction.
Let me demonstrate this explicitly. Suppose you have an ideal solenoid of radius $r_0$, with $n$ turns per unit length, oriented along the $z$ axis. Its magnetic field is given by
$$\vec{B} = \begin{cases}\mu_0 n I\hat{z} & r < r_0 \\ 0 & r > r_0\end{cases}$$
As you've noticed, this implies that $\nabla\times\vec{E} = 0$ outside the solenoid. Now, you might think that implies the integral $\oint\vec{E}\cdot\mathrm{d}\ell$ around a loop outside the solenoid, which gives the EMF, must be zero. But that's not actually the case. The relationship between $\nabla\times\vec{E}$ and $\oint\vec{E}\cdot\mathrm{d}\ell$ comes from Stokes' theorem, and it says
$$\oint_{\mathcal{C}}\vec{E}\cdot\mathrm{d}\ell = \iint_{\mathcal{S}}(\nabla\times\vec{E})\cdot\mathrm{d}^2\vec{A}$$
So the line integral around the loop is determined by the curl of $\vec{E}$ everywhere inside the loop, including inside the solenoid where
$$\nabla\times\vec{E} = -\mu_0 n \frac{\partial I}{\partial t}\hat{z}\quad(r < r_0)$$
Performing the integral gives you
$$\mathcal{E} = \oint_{\mathcal{C}}\vec{E}\cdot\mathrm{d}\ell = \iint_{\mathcal{S}}(\nabla\times\vec{E})\cdot\mathrm{d}^2\vec{A} = -\int_0^{2\pi}\int_{0}^{r_0}\mu_0 n \frac{\partial I}{\partial t}\hat{z}\cdot r\mathrm{d}r\,\mathrm{d}\theta\,\hat{z} = -\mu_0 \pi r_0^2 n\frac{\partial I}{\partial t}$$
so you can see that any time-varying current in the solenoid will create an EMF and induce a current.
Best Answer
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.