What is the evidence of interpreting $g_{\mu\nu}$ as the metric of space-time? I think if we don't mention the meaning of $g_{\mu\nu}$ as the metric of space-time, we can still construct the equation of motion and Einstein field equation in a way such that $g_{\mu\nu}$ is just a tensor field in a "flat" (independent) space-time.
What is the reason that $g_{\mu\nu}$ must be the geometric of space-time (according to experiments or self-consistency of the theory,...) besides the fact that it is an elegant interpretation?
 A: I don't think there is one. In fact, I know for sure that approaches has been proposed which treat $g_{\mu \nu}$ as a physical field, which can be interpreted as the metric in some effective Riemannian manifold.
Take a look at the relativistic theory of gravitation by Logunov. It attempts to solve the problem of energy conservation by imposing that $g_{\mu \nu}$ is not geometrical, but rather physical.
But this approach is not widely accepted, and there is a good reason for that: elegance of General Relativity and its great predictive power.
A: if gμν is a tensor field, then as per experimental verification it has to interact in the same way as photon as well as massive objects.interaction does not depend on other objects only on source objects that are causing stress energy tensor.
another evidence is gravitational waves if detected. then we can see the difference more easily. because at the LIGO facility , two beams of lights are bouncing in perpendicular direction , if gravitational waves strong enough sensed by ligo detector that can change the distance between arms of detector.so gravitational waves is changing the spacetime between them physically. see the effects of gravitational waves.
so we can say it is the metric of manifold not a field. 
A: I think the key point to understand is that metrics  exist which describe a flat space-time and others which describe a curved space. Both can be distinguished by the Riemann's curvature tensor which is zero for a metric which belongs to a flat space-time whereas for a curved space it is non-zero.
Therefore if the curvature tensor is evaluated to be non-zero, also the "surrounding" space-time must be curved. I cannot imagine it otherwise.
Example: do a coordinate tramsformation $x=x' cos(\Omega t) - y' sin(\Omega t)$; $y=x' cos(\Omega t) + y' sin(\Omega t)$ $z'=z$ on $ds^2 =dt^2-dx^2-dy^2-dz^2$ One gets a non-diagonal metric which belongs to a flat space-time. Compute the Riemann tensor to check it out (I admit : a lot of work). But I think the idea is suggestive enough to be clear.
2.) the Schwarzschild metric. This metric cannot be changed to a flat metric. It is really associated to a curved space. There is no coordinate transformation which could make the space-time described by a Schwarzschild metric everywhere flat (at most it could be "flat-like" at one single point ... in fact for real flatness several points are needed with flat-like behaviour). 
Again, I cannot imagine to disconnect $g_{\mu\nu}$ from its original function  $ds^2 =g_{\mu\nu}dx^{\mu}dx^{\nu}$. I could attribute to $g_{\mu\nu}$ everything what I would want except $ds^2 =g_{\mu\nu}dx^{\mu}dx^{\nu}$.
However, matrix elements of $g_{\mu\nu}$ were already measured for instance by measuring the time dilation of clocks at different places in 
a gravitational field or by the light deviation in the metric of the sun. So the metric is real as an electromagnetical field is. 
