Does general relativity (GR) solve mass-equivalence problem or just renames it? It is usually said that GR solved problem of equivalence of gravitational and inertial mass. Yet, my objection is:
OK, there is no gravitational mass in GR, but Einstein had to introduce mass as "spacetime curving constant".
So the question A: Why is the gravitational and inertial mass the same?
is replaced by the question:
B: Why is inertial mass and spacetime-curving constant the same?
I am honestly unable to clearly solve my objection, I have troubles thinking abstract enough. Is or is not my objection correct?
Is is possible to imagine a universe, where, for example, electron has two different masses: one (inertial) appears in EM interactions (and other non-gravitational interactions), other appears in spacetime-curving equations.
A: The "why" questions about physical laws are rarely questions that can actually be answered by physics. I can just tell you that Einstein gravity was constructed so that there is no difference between inertial and gravitational mass. You can see this from the Einstein field equations which read
$$R^{\mu\nu} - \frac{1}{2} R g^{\mu\nu} = \frac{8 \pi G}{c^2} T^{\mu\nu}$$
where the left hand side corresponds to the geometry of the space-time, and the right-hand side to the matter-energy sources. 
For instance, for a coherent cloud of dust, we would just have $T^{\mu\nu} = \rho u^\mu u^\nu$, where $\rho$ is the matter density and $u^\mu$ the four-velocity field. (Roughly speaking, in the Newtonian limit only the $\mu\nu=00$ component is relevant and $\rho u^0 u^0$ is the matter density.) So the field equations tell you "space-time geometry (gravitational field) is sourced by matter-energy".
But now consider the following. The left-hand side automatically fulfills
$$\left(R^{\mu\nu} - \frac{1}{2} R g^{\mu\nu}\right)_{;\nu} = 0$$
where $;\nu$ means a space-time-covariant gradient and I use the Einstein summation convention. That is, the covariant divergence of the left-hand side of the Einstein equations automatically vanishes. So, if the Einstein equations are fulfilled, this also implies
$$T^{\mu\nu}_{\;\;\;;\nu} = 0$$
If you aply this to dust and combine it with the conservation of matter (continuity equation) $(\rho u^\mu)_{;\mu} = 0$, you just get the geodesic equation $u^\mu_{;\nu}u^\nu = 0$! (In the Newtonian limit this just gives you that all particles experience a universal gravitational acceleration $\vec{a}$.)
In other words, the fact that the matter sources the geometry implies the equations of motion and the universal gravitational acceleration! It is not possible to separate any "space-time curving" constant as compared to an "inertial mass" in relativity. So you do not even have the option to set them as different. 
This is actually true for all metric gravities with "minimal coupling" of matter to the geometry. However, there certainly are modified gravities where different types of matter may couple to slightly different geometries. This cannot be quite characterized as inertial and gravitational mass being non-equal, but it certainly does violate the universality of gravitational acceleration (the weak equivalence principle). Once again, I cannot quite tell you "why" this or that theory should apply, I could only invoke some kind of "esthetic" or "simplicity" criterion.
