The London equation almost follows from the Drude model of conductivity and Maxwell's equations. Here's how.
In the Drude model, we assume that electrons moving through a metal are subject to two interactions. First, they experience a certain force $\vec{F}$, and accelerate in response to it. Second, there is a probability of an electron hitting a nucleus. When this happens, we assume that it stops dead. In an amount of time $\Delta t$, we assume that this probability is approximately $\Delta t/\tau$.
Based on these assumptions, it is not too hard to show that the total momentum $\vec{p}$ of a bunch of charge carriers in some volume of a metal is governed by the equation
$$
\frac{d \vec{p}}{dt} = - \frac{1}{\tau} \vec{p} + \vec{F}.
$$
But the momentum of these charge carriers is proportional to the current density in the metal ($\vec{J} \propto \vec{p}$), and the force on them is proportional to the electric field ($\vec{E} \propto \vec{F}$). If we put in all the appropriate constants, this equation can be rearranged to become
$$
\frac{\partial \vec{J}}{\partial t} = - \frac{1}{\tau} \vec{J} + \frac{n q^2}{m} \vec{E},
$$
where $n$, $q$, and $m$ are the number density, charge, and mass of the charge carriers (respectively). For a steady-state current ($\dot{\vec{J}} = 0$), this implies that
$$
\vec{J} = \frac{n q^2 \tau}{m} \vec{E},
$$
which can be recognized as the microscopic version of Ohm's Law, with $\sigma = n q^2 \tau/m$.
A perfect conductor, however, is one where $\tau \to \infty$. This implies that
$$
\frac{\partial \vec{J}}{\partial t} = \frac{n q^2}{m} \vec{E},
$$
If we take the curl of this equation, and apply Faraday's Law, we have
$$
\frac{\partial}{\partial t} (\vec{\nabla} \times \vec{J}) = \frac{n q^2}{m} \vec{\nabla} \times \vec{E} = - \frac{n q^2}{m} \frac{\partial \vec{B}}{\partial t}.
$$
Thus, we must have
$$
\vec{\nabla} \times \vec{J} + \frac{n q^2}{m} \vec{B} = \text{constant with respect to $t$.}
$$
So far we've used nothing but the Drude model and Maxwell's equations. The fundamental assumption of the London equation (and the part that does not follow from Maxwell's equations) is that the constant in this last equation is exactly zero:
$$
\vec{\nabla} \times \vec{J} + \frac{n q^2}{m} \vec{B} = 0.
$$
Once we have this, we can then show that $\vec{B}$ nearly vanishes inside the bulk of a superconductor. If we take the curl of Ampere's Law, and assume static fields, it's not too hard to show that we have
$$
\nabla^2 \vec{B} = \frac{\mu_0 n q^2}{m} \vec{B}
$$
which implies that $\vec{B}$ must either die off or grow exponentially inside a superconductor. Growing exponentially is right out, so it must be the case that magnetic fields decay as you move deeper into a superconductor. The characteristic length scale for this die-off is $\lambda = \sqrt{m/\mu_0 n q^2}$, and this works out to be on the order of a few dozen nanometers for most superconductors.
