This is a weird way of phrasing, but it seems about right. The interior metric for a Schwarzschild black hole can be written as
$$\mathrm{d}s^2=-\left(1-\frac{2M}{r}\right)\mathrm{d}t^2+\left(1-\frac{2M}{r}\right)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2,$$
where we are using Schwarzschild coordinates. In these coordinates, the singularity is located at $r=0$. However, there is something odd: since $r<2M$, the $t$ coordinate is spacelike and the $r$ coordinate is timelike, so they don't have their usual interpretations.
To make the expression look more natural, we can then write
$$\mathrm{d}s^2=\left(\frac{2M}{r}-1\right)\mathrm{d}t^2-\left(\frac{2M}{r}-1\right)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2,$$
which doesn't change anything. The metric is still the same, and so are the coordinates.
Since we usually like to use the notation $t$ for a time coordinate and $r$ for a spatial coordinate, let us define new coordinates $t'$ and $r'$ by
$$t = r' \quad \text{and} \quad r = t'.$$
Using these coordinates, the metric becomes
$$\mathrm{d}s^2=\left(\frac{2M}{t'}-1\right)\mathrm{d}r'^2-\left(\frac{2M}{t'}-1\right)^{-1}\mathrm{d}t'^2+t'^2\mathrm{d}\Omega^2.$$
We usually write the time coordinate first, so we finally get
$$\mathrm{d}s^2=-\left(\frac{2M}{t'}-1\right)^{-1}\mathrm{d}t'^2+\left(\frac{2M}{t'}-1\right)\mathrm{d}r'^2+t'^2\mathrm{d}\Omega^2.$$
Notice that Birkhoff's theorem ensures the only possible solution is the Schwarzschild solution, since the spacetime is in vacuum and it is spherically symmetric.
Alternative Derivation Using Interior Solutions
Alternatively, we can obtain the solution using interior solutions. In this case, the solution is the Tolman–Oppenheimer–Volkoff solution. We take the stress tensor to be a perfect fluid and the equations are
\begin{gather}
\mathrm{d}s^2=-e^{2\phi(r)}\mathrm{d}t^2+\left(1-\frac{2m(r)}{r}\right)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2, \\
\frac{\mathrm{d} m}{\mathrm{d} r} = 4 \pi \rho(r) r^2, \\
\frac{\mathrm{d} P}{\mathrm{d} r} = - (P(r) + \rho(r))\frac{4 \pi P(r) r^3 + m(r)}{r[r-2m(r)]}, \\
\frac{\mathrm{d} \phi}{\mathrm{d} r} = \frac{4 \pi P(r) r^3 + m(r)}{r[r-2m(r)]}.
\end{gather}
Let us solve each of these equations one by one. Since the interior is made of vacuum, we know already that $\rho(r) = P(r) = 0$. First we solve the equation for $m$. At the boundary $r=R$ we need to have $m(R) = M$, but the equation for $m$ is simply
$$\frac{\mathrm{d} m}{\mathrm{d} r} = 0,$$
since $\rho = 0$. Hence, we conclude $m(r) = M$ for all $r$. This is unusual and singular at the origin, but it is the only possible solution that matches the exterior solution.
The equation for pressure is trivial.
The equation for $\phi$ then becomes
$$\frac{\mathrm{d} \phi}{\mathrm{d} r} = \frac{M}{r[r-2M]},$$
which can be solved with the condition $e^{2\phi(R)} = \frac{1}{2}\log\left(1 - \frac{2M}{R}\right)$ (necessary to match the exterior Schwarzschild solution) to yield
$$e^{2\phi(r)} = 1 - \frac{2M}{r}.$$
Plugging these expressions for $m$ and $\phi$ in the TOV metric, you get back
$$\mathrm{d}s^2=-\left(1-\frac{2M}{r}\right)\mathrm{d}t^2+\left(1-\frac{2M}{r}\right)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\Omega^2,$$
as expected. And then you can proceed with the argument in the beginning of the question.
There is a caveat: technically these coordinates are awful to study anything very close to the event horizon, and notice the expressions would have led to terrible results if I had used the correct value of $R = 2M$. However, you can understand that this derivation assumes we are working only outside of the black hole, or matching to a Schwarzschild solution that holds down to $R = 2M - \epsilon$ or something. This is an important caveat, and a better derivation for this result is definitely Birkhoff's theorem.
