Yes, it absolutely is an exact coordinate transformation!
Take the expression above, and write it as a line element (also, take $\epsilon$ to be either plus or minus 1 to avoid writing $\pm$ over and over), and take $d\Omega^{2} = \left(d\theta^{2} + \sin^{2}\theta\,d\phi^{2}\right)$:
$$ds^{2} = -\left(1-\frac{2M}{r}\right)dt^{2} + 2dt dr\left( -\frac{2M\epsilon}{r}\right) + \left(1+\frac{2M}{r}\right)dr^{2} + r^{2}d\Omega^{2}$$
By inspection of this expression, we can see that the $g_{tt}$ element is already in schwarzschild form, and the angular parts are already correct. We just need the off-diagnonal elements gone, and we need a different $g_{rr}$. with all of these constraints, we guess that we cannot do very much, and are pretty much restricted to a coordinate change of the form
$$t = T + f(r)$$
where we expect $T$ to be the Schwarzschild time. Substituting this ansatz into our line element, we have:
$$ds^{2} = -\left(1-\frac{2M}{r}\right)dT^{2} + 2dTdr\left(-f^{\prime}\left(1-\frac{2M}{r}\right) -\frac{2M\epsilon}{r}\right) + \left(-f^{\prime 2}\left(1-\frac{2M}{r}\right) -\frac{4M\epsilon f^{\prime}}{r}+ 1+\frac{2M}{r}\right)dr^{2} + r^{2}d\Omega^{2}$$
The diagonal term vanishes if we have:
$$f^{\prime} = -\frac{\epsilon \frac{2M}{r}}{1-\frac{2M}{r}}$$
so, a convenient choice for $f$ is $-2M\epsilon {\rm ln}\left(\frac{r}{2M} -1\right)$
Finally, all we have to do is check to see whether our expression for $g_{rr}$ checks. Substituting the value for $f^{\prime}$, we have:
$$g_{rr} = -\left(-\frac{\frac{2M \epsilon}{r}}{1-\frac{2M}{r}}\right)^{2}\left(1-\frac{2M}{r}\right) -\frac{4M\epsilon}{r}\left(-\frac{\epsilon \frac{2M}{r}}{1-\frac{2M}{r}}\right)+ 1+\frac{2M}{r}$$
The second term is $-2$ times the first term, and multiplying the last term by $\frac{1-2M/r}{1-2M/r}$ and adding everything together (and using $\epsilon^2=1$), we find that we do, indeed end up with $\frac{1}{1-2M/r}$, and we are done. $T$ is indeed the schwarzchild coordinate time.
