Schwarzschild solution, stress-energy side of Einstein's equation

Your question is understandable: if the metric is curved by a mass, shouldn't there be a mass in the center, like there is a charge in the center of an electrostatics problem?
But the answer is no, because there is no mass. Let me explain.

In deriving the Schwarzschild solution, we impose that, in the Einstein Equations (EE), $T^{\mu\nu}=0$: it's a vacuum, spherically symmetric, solution (it's the only one actually, according to Birkhoff's theorem). We then impose the symmetries to the metric tensor, and plug it into our vacuum EE to find that there is only one free parameter, $m_0$, the Misner-Sharp """mass""", that is completely invariant
$\frac{\partial}{\partial t}m_0=\frac{\partial}{\partial r}m_0=0$
and that has a form of
$1-\frac{2m_0}{R}=-g^{\mu\nu}\partial_\mu R \partial_\nu R$
where $R$ is the coefficient of the angular part of the metric. The Schwarzschild metric then takes the form
$ds^2=\left(1-\frac{2m_0}{r}\right)dt^2-\left(1-\frac{2m_0}{r}\right)^{-1}dr^2-r^2\left(d\theta^2+sin^2\theta d\phi^2\right)$.
Now one would like to find a physical meaning for $m_0$, and by comparison with the Newtonian limit, $g_{00}\simeq1+\frac{2}{c^2}\frac{-GM}{r}$ where $M$ is the mass of the Newtonian source. In God-given-units $(c=G=1)$ one can see that $M$ is the Newtonian interpretation of $m_0$. This is the only reason why we say that $m_0$ is a mass: if we don't take Newton into account, it is only a parameter in the metric.

Edit:
As I said in the comments, one could decide to study a different, modified Schwarzschild metric, where one imposes $T^{\mu\nu}\sim m\delta (\vec{r})$: there is an interesting paper by Fiziev where this alternative is considered. Please note: this isn't the mathematical ideal framework that we call Schwarzschild metric, so the answer to your question would still be no, there is no Dirac-delta in the stress-energy tensor of the Schwarzschild solution.

The Schwarzschild solution is a vacuum solution. There is no mass in it anywhere. If you are using it to model e.g. a planet then it only applies in the vacuum region outside of the planet. For this reason, I prefer to think of the free parameter in the Schwarzschild spacetime as a length (the Schwarzschild radius) rather than a mass.

In the case of a black hole, the coordinate r=0 with the singularity must be removed from the manifold. The infinite curvature at the singularity ruins the manifold structure. So you cannot put a delta function at r=0 since r=0 is not a part of the manifold.

For the case of a spherically symmetric planet or star, the Schwarzschild spacetime only describes the vacuum region outside the mass, and inside the mass you must use an appropriate interior Schwarzschild solution. The vacuum solution is unique, but the interior solution will depend on the details of the stress-energy distribution

Singularities in GR are little more complicated than just putting some distribution somewhere in the equations. In Maxwell theory, the singularity has no influence on underlying structure. In GR, singularity means problems with underlying structure itself. Singularity in GR literally means some border, where geodesics end in finite amount of affine parameter and yet it is mathematically impossible to extend the spacetime to make the geodesics continue. It is border where the spacetime itself ends.

The distribution like dirac delta function is defined by its behavior under integration. In Maxwell theory you can write and compute charge $q=\int\rho(\vec{x})dV,$ even if the charge distribution $\rho$ is delta function, because the space over which you are integrating is still well-behaved. This makes dirac delta function useful concept. Not so in general relativity. In GR there is problem in volume integral itself, not just the integrand. The integral is defined only on spacetime, but the singular point you would like to describe by some generalized function does not belong to spacetime. The integral is simply undefined and you would really need to generalize the notion of integral (or manifold) itself. I do not have any knowledge that such generalization exists.

Edit:

Just to describe the problem more closely, imagine you would wish to get rid of the problem of singular point not being in spacetime by gluing the singular border together. Simply put, you would need to decide how will the geodesic that ends in singularity emerges at the other side. Now take for example timelike geodesic. In Schwarschild spacetime, this would mean that ingoing particle will start to move away from singularity. This means, it would either become spacelike, or it would start to travel back in time. So you see, the whole idea of continuation through singularity is quite problematic. The best you could do, is that the geodesic will remain at singularity, but the time for it will go on. But because the point is singular, this makes no sense. What is the notion of time in singularity itself? There is no such thing, you would need to invent it. And then you would be stuck with trying to build your concepts inside the singular point. Schwarzschild spacetime is simply irremediable without extensive (and unnatural) modifications.

But there are "distribution-like" singularities in GR like for example conical singularity. The spacetime for conical singularity is everywhere flat except the fact that there is some central (spacelike) line around which angle is not $2\pi$, but some different number. How will the infalling geodesic continue after reaching the central line? Sensible choice is to just make a straight line in your cartesian coordinate chart. There is no problem doing this. And if you do, you can use Einstein field equations and you will get that there is some distribution in stress-energy in central line. It will make sense. But you cannot do this for Schwarzschild, there the singularity is much more serious.