The short answer is that the Einstein field equation relates the energy, momentum, and stress of matter to curvature in such a way so that it would be true. We could certainly imagine other relationships between matter and curvature in which the opposite would be the case, or some other stranger things; they just don't resemble how our universe works.
I will provide an overview of some facts regarding differential geometry and how they connect to this question. Depending on your background, this may be insufficient as a full explanation, but at least it will provide you with pointers about what to follow up on (although most of it would be more relevant either in math.SE or physics.SE, depending).
On a (pseudo-)Riemannian manifold, a curve $gamma$ is an affine geodesic whenever it can be parametrized in such a way as to keep the same velocity vector $u = dot{gamma}$, or slightly more formally, whenever its velocity vector parallel-transported along the curve remains the same. Equivalently, one can think of affine geodesics are curves of zero acceleration along themselves, which is expressed by the geodesic equation:
$$nabla_uu = 0text{.}$$
There is an obvious intuitive connection between this and inertial motion, so we can think of these geodesics as describing the motion of free test particles. Unless the length along the geodesic is zero (which can happen on a pseudo-Riemannian manifold), we can take the length along it to be the affine parametrization. Later in physics, the length along the worldline of a massive particle is the time measured by a clock going along it--its proper time $tau$.
If you have you have two nearby geodesics with the same velocity $u$ but separated by a vector $n$ (i.e., this vector connects points of equal affine parameter), we can talk about the relative acceleration between those geodesics. This is known as the geodesic deviation equation, and in terms of the Riemann curvature $R$, one of its forms would be:
$$nabla^2_{u,u}n + R(n,u)u = 0text{,}$$
or equivalently in coordinates:
$$frac{mathrm{D}^2n^alpha}{mathrm{d}tau^2} + underbrace{R^alpha{}_{mubetanu}u^mu u^nu}_{mathbb{E}^alpha{}_beta},x^beta = 0text{,}$$
where the tidal tensor $mathbb{E}^alpha{}_beta$ is sometimes called the 'electric' part of the Riemann curvature tensor.
If you're not familiar with how tensors work, for now just imagine the Riemann tensor as a mathematical machine that takes three vectors and spits out another one in a certain way. In four dimensions, the tensor has $256$ components, although only $20$ of them are algebraically independent.
Anyway, once you get used to the idea that nearby geodesics can deviate from one another due to the presence of curvature, we can ask a similar question: suppose you have a small ball (size $epsilon$) of test particles, all initially comoving with the same velocity $u$. Since their geodesics will have some relative acceleration, the ball generally won't stay of the same volume (or stay a ball, for that matter). In an inertial frame comoving with the test particles, we can consider the initial fraction acceleration of its volume in the limit of zero size of the ball:
$$lim_{epsilonto 0}left.frac{ddot{V}}{V}right|_{t=0}!= -underbrace{R^alpha{}_{mualphanu}}_{R_{munu}} u^mu u^nu = -mathbb{E}^alpha{}_{alpha}text{,}$$
where the Ricci curvature $R_{munu}$ is the trace of the Riemann curvature, and the whole expression is equivalent to the trace of the tidal tensor. A more general situation is described by the Raychaudhuri equation.
The takeaway is this: the Ricci curvature tells us whether small balls of initially comoving test particles will expand or contract, i.e. move towards or away from each other. (NB: There's almost no physics in this section--other than borrowing some vocabulary to help frame these issues in a way that will more readily connect to physics, these are facts of differential geometry.)
To have physical content, we need to not only interpret spacetime as a four-dimensional manifold, but also say something about what the curvature is in terms of physical quantities. Given the stress-energy-momentum tensor $T_{munu}$ and its trace $T$, the Einstein field equation provides the Ricci curvature:
$$R_{munu} = kappaleft(T_{munu} - frac{1}{2}g_{munu}Tright)tag{EFE}text{,}$$
where $kappa = 8pi G/c^4$ is a constant. So does that mean that matter attracts or not? Well, it still depends: locally, we can say that under gravitational freefall, a small ball of initially comoving test particles with initial four-velocity $u$ will contract whenever $R_{munu}u^mu u^nugeq 0$. Substituting in the above Einstein field equation, we get the strong energy condition, which basically boils down to 'gravity is always locally attractive'.
However, even though the strong energy condition might not necessary hold (and gravity can be repulsive, e.g. on the cosmological scale due to dark energy), it holds under ordinary circumstances. That's because in the local inertial frame comoving with $u$, the trace of the stress-energy tensor is $T = -rho + 3p$, where $p$ is the pressure (pretending for simplicity that we're dealing with a perfect fluid; otherwise, we can take average of the principal stresses), while $rho = T_{munu}u^mu u^nu$ is the energy density. Therefore, the time-time projection of the Einstein field equation is (in units of $G=c=1$):
$$mathbb{E}^alpha{}_alpha equiv R_{munu} u^mu u^nu = 4pi(rho+3p)text{.}$$
Since energy density and pressure have equivalent units, to compare, ordinary water has a density of $(1;mathrm{kg}/mathrm{L})c^2 = 9.0times 10^{19},mathrm{Pa} = 8.9times 10^{14},mathrm{atm}$. Ordinary matter has overwhelmingly higher energy density than stress.
The above time-time projection of the Einstein field equation is the direct analogue of the Gauss' law for Newtonian gravity (matter density) or electromagnetism (charge density). In the weak-field limit (small deviation from flatness), we can use this to recover Poisson's equation for Newtonian gravity; see also part of this answer.
In other words, the Einstein field equations guarantees that not only is ordinary matter locally attractive, but that in the weak-field limit, it is essentially Newtonian as well.
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