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One very important plus for the path integral formulation is that it generalizes directly to quantum fields, or rather applications there, so in that sense it is mathematically nicer than the Schrödinger formulation. Additionally, even without fields, path integrals lend themselves very well to proving certain features of the physical world, some of which don't even require any explicit calculation to derive.

As to your question about the physical basis of the Schrödinger equation, I wish that I've read his original paper, because undoubtedly his motivations are different. Still, let me ramble on a bit and see if it is sufficient to at least partially sate your curiosity. The only non-classical physics assumed here is that momentum and position do not commute.

First, some hand-waving, using some properties a propagator U(t,t_0) "should" have. First, it must conserve probability, which means it must be unitary, as well as being continuous (specifically, reduce to identity in the limit t->t_0). Second, it must have the composition property U(t,t_0) = U(t,t_1)U(t_1,t_0). For any Hermitian operator H, U(t+dt,t) = 1-iH dt satisfies these conditions. Now, here we increase our hand-waving amplitude and state that by analogy with classical mechanics, where the time-evolution of observables is determined by the Hamiltonian, H should not be just any Hermitian operator, but specifically the Hamiltonian (for consistency in turning it into an operator, have a factor of 1/ℏ = 1/hbar). From there, the time-composition property gives
U(t+dt,t_0) - U(t,t_0) = -iH/ℏ dt U(t,t_0),
which directly gives the time-dependent Schrödinger equation for the propagator. From there, it is at least not "too" difficult to derive the Schrödinger equation for the wavefunction or other versions.

Second, let's make clear(er) the connection to classical mechanics. Since you've read Landau and Lifschitz, I assume you're already familiar with most of what follows, so I'll be very brief. By 'functions', here I'll mean 'smooth functions' to remove various complications in which that is not the case. As you say, Shankar simply states the Schrödinger equation as corresponding to the Hamilton equations of classical mechanics. This is actually not just an analogy of roles--the Hamilton equations and the Schrödinger equation are connected through a direct deformation to make p,q non-commutative. Of course, Shankar does not say this at all, but the details can be filled.

Given a manifold M, a Lagrangian is a function on its tangent bundle and time, L:TM×R->R. From the principle of stationary action, we get the Euler-Lagrange equations
(d/dt)(p_k) = F_k,
where p_k = ∂L/∂q'^k being momentum and F_k = ∂L/∂q^k the force corresponding to generalized coordinate pair (q^k,q'^k). The energy function is E(q,q',t) = p_kq'^k - L, and in a closed system (Lagrangian not explicitly dependent on time), this is conserved. Formally, the Hamiltonian is found via a Legendre transformation λ(q,q') = (p,q). Explicitly, the Hamiltonian a function on the cotangent bundle ("phase space") and time (such functions are observables), H:T*M×R->R, such that H composed with λ is E. The Legendre transformation applied to the Euler-Lagrange equations give the Hamilton equations (implicit assumption: λ is a diffeomorphism):
q'^k = ∂H/∂p_k, p'_k = -∂H/∂q_k, ∂H/∂t = -∂L/∂t.
Given the Poisson bracket of any two observables, {f,g} = (∂f/∂p_k)(∂g/∂q^k) - (∂f/∂q^k)(∂g/∂p_k), one can write the time-evolution of any observable on a path that solves the Hamilton equation as
df/dt = ∂f/∂t + {H,f}.
This explains the "time-evolution of observables is determined by the Hamiltonian" comment above. Typically, we assume that ∂f/∂t = 0 because observables explicitly dependent on time complicate things unnecessarily without introducing any interesting physics. Both the momentum components and the position components are classical observables defining a classical state (which is, after all, just the position in the phase space). So, in classical mechanics, we have a Hamiltonian vector field {H,·} on the classical phase space, i.e.,
The time-evolution of a classical state is given by the Hamiltonian operator (vector field on the cotangent bundle) acting on the state.
This is nothing more than the Hamilton equations re-packaged.

Formally, the algebra of observables is a commutative C^*-algebra which after deformation [q,p] = iℏ is a non-commutative C^*-algebra that is *-isomorphic to the algebra of bounded operators on some Hilbert space, and since the Hamiltonian is also itself an observable, we should have:
The time-evolution of a quantum state is given by the Hamiltonian operator (linear operator on the Hilbert space) acting on the state.
Here, forgive me for hand-waving away the mathematical details, but this should make the following completely unsurprising:
iℏ d|ψ>/dt = H|ψ>,
where |ψ> is a quantum state vector (ket) and H is the Hamiltonian. This is just another version of the Schrödinger equation.

Edited for clarity, to make the connection more explicit.