In the context of axiomatic quantum field theory, there is a theorem (see theorem 3-7 in PCT, Spin and Statistics, and All That by Streater and Wightman, who I will refer to as "SW"), which SW call the "reconstruction theorem," essentially stating that correlation functions serve to completely determine a corresponding field theory in the Hilbert Space formalism. Specifically, they show that (I paraphrase for brevity)
Given a sequence $\mathscr W^{(n)}$ of tempered distributions defined by $n$ spacetime points (correlation functions) that satisfy certain technical properties (cluster decomposition, relativistic transformation law, etc.) there exists a separable Hilbert space $\mathscr H$, a continuous unitary representation $U$ of $\mathbb R^{3,1}\rtimes \mathrm{SO}^+(3,1)$ (the proper, orthochronus Poincare group) on $\mathscr H$, a unique Poincare-invariant vacuum state $|0\rangle$, and a hermitian scalar field $\phi$ with appropriate domain such that
\begin{align}
\langle 0|\phi(x_1)\cdots \phi(x_n)|0\rangle = \mathscr W^{(n)}(x_1, \dots, x_n)
\end{align}
Furthermore, any other field theory with these vacuum expectation values (vevs) is unitary equivalent to this one.
In other words, vevs determine a field theory up to unitary equivalence, and a sequence of sufficiently well-behaved correlation functions completely determines a field theory with given vevs, so correlators determine a field theory up to unitary equivalence.
The Upshot. Since the path integral allows you to compute all correlators in principle through the (somewhat schematic) formula
\begin{align}
\langle \phi(x_1)\cdots \phi(x_n)\rangle = \frac{\int [d\phi] \phi(x_1)\cdots \phi(x_n)e^{i S[\phi]}}{\int [d\phi] e^{i S[\phi]} }
\end{align}
the path integral gives a complete characterization of a given field theory.
Note. I am by no means an expert on axiomatic quantum field theory, so if I have said anything here that isn't strictly, mathematically correct, I apologize ahead of time. Also, I'm not certain how general SW's characterization of field theory is, so my remarks are not completely general, but I would think that the spirit of these remarks is thought to hold for all (or most) physical quantum field theories.
Also, this is certainly not a particularly physical answer. I'd be curious to hear from another user about the physical intuition behind why one might expect correlators to be so fundamental and all-encompassing.
the actual non-relativistic part is the Schrödinger equation
Indeed. Hence, people tried to come up with a Lorentz-invariant evolution equation for the wave function such as the Klein-Gordan and Dirac equations (as mentioned by anna v). However, interpretation of solutions of these equations as probability densities are problematic (in case of the Klein-Gordon equation, the predicted probabilities won't be positive definite - though the Dirac equation 'fixes' this particular problem).
We've come to realize that these equations are not to be understood as evolution equations for the quantum state, but in the sense of quantum field theory, where (in the Schrödinger picture), their (classical) solutions correspond to the configurations of the system (ie the equivalent of positions in single-particle QM), with states being wave-functionals on that solution space.
Best Answer
It is not quite true that we don't obtain any useful information. The relativistic particle action is indeed $$ S = -m_0c^2 \int dt_{\rm proper} $$ When you substitute your correct formula for $dt_{\rm proper}$ and Taylor expand the Lorentz factor in it, the integral has the factor of $dt_{\rm coordinate}(1-v^2/c^2+\dots)$. The first term proportional to $1$ is constant and the second term gives you the usual $mv^2/2$ part of the non-relativistic action.
At least formally, the relativistic action above may be used to deduce the propagators for a relativistic spinless particle – such as the Higgs boson. The addition of a spin isn't straightforward. However, the "proper time" action above may be easily generalized to the "proper area of a world sheet" action (times $-T$, the negative string tension) for string theory, the so-called Nambu-Goto action, and this action admits spin, interactions, many strings/particles, and is fully consistent. The "proper time" action is therefore the usual starting point to motivate the stringy actions (see e.g. Polchinski's String Theory, initial chapters).
The reason why a single relativistic particle isn't consistent without the whole machinery of quantum fields is a physical one and it may be seen in the operator formalism just like in the path integral formalism. Any valid formalism has to give the right answers to physical questions and the only right answer to the question whether a theory of interacting relativistic particles without particle production may be consistent is No.
In the path integral formalism, we could say that it brings extra subtleties to have a path integral with a square root such as $\sqrt{1-v^2/c^2}$. To know how to integrate such nonlinear functions in an infinite-dimensional functional integral, you have to do some substitutions to convert them to a Gaussian i.e. $\exp(-X^2)$ path integral.
This may be done by the introduction of an auxiliary time-like parameter along the world line, $\tau$, agreeing with the variable in the aforementioned paper. With a condition relating $\tau$ and $t_{\rm coordinate}$, it may be guaranteed that the new action in the $\tau$ language looks like $$ - m\int d\tau \,e(\tau) \left( \frac{d X^\mu}{d\tau}\cdot \frac{dX_\mu}{d\tau}\right) $$ which is nicely bilinear and the square root disappear. However, this clever substitution or any similar substitution has the effect of allowing the negative-energy solutions, too.
While the relativistic $p^2/2m$ is positive semidefinite, $m/\sqrt{1-v^2/c^2}$ can really have both signs. We may manually try to "forbid" the negative sign of the square root but this solution will always reappear whenever we try to define the path integral (or another piece of formalism) rigorously.
This implies that we have states with energy unbounded from below, an instability of the theory because the particle may roll down to minus infinity in energy. Alternatively, the squared norms of these negative-energy states may be (and, in fact, should be) taken to be negative, traded for the negative energy, which brings an even worse inconsistency: negative probabilities.
The only consistent way to deal with these negative-norm solutions is to "occupy" all the states with negative energies so that any change in the states with negative energy means to add a hole – an antiparticle such as the positron – whose energy is positive (above the physical vacuum) again. At least, this description (the "Dirac sea") is valid for fermions. For bosons, we use an approach that is the direct mathematical counterpart of the Dirac sea but only in some other variables.
It's important to realize that any attempt to ban the negative-energy solutions by hand will lead to an inconsistent theory. The consistent theory has to allow the antiparticles (which may be identical to the particles in some "totally real field" cases, however), and it must allow the particle-antiparticle pairs to be created and destroyed. It's really an inevitable consequence of the combination of assumptions "special relativity" plus "quantum mechanics". Quantum field theory is the class of minimal theories that obey both sets of principles; string theory is a bit more general one (and the only other known, aside from QFT, that does solve the constraints consistently).
Why quantum field theory predicts a theory equivalent to multibody relativistic particles (which are indistinguishable) is the #1 most basic derivation in each quantum field theory course. A quantum field is an infinite-dimensional harmonic oscillator and each raising operator $a^\dagger(\vec k)$ increases the energy (eigenvalue of the free Hamiltonian $H$) by $\hbar\omega$ which is calculable and when you calculate it, you simply get $+\sqrt{m_0^2+|\vec k|^2}$ in the $c=1$ units. So $a^\dagger(\vec k_1)\cdots a^\dagger(\vec k_n)|0\rangle$ may be identified with the basis vector $|\vec k_1,\dots,\vec k_n\rangle$ (anti)symmetrized over the momenta (with the right normalization factor added) in the usual multiparticle quantum mechanics. This has various aspects etc. that are taught in basic quantum field theory courses. If you don't understand something about those things, you should probably ask a more specific question about some step you don't understand. Quantum field theory courses often occupy several semesters so it's unproductive to try to preemptively answer every question you could have.