The boundary of your solid torus $T_L \cong D^2 \times S^1$ is equipped with a pair of foliations by circles (at "right angles"). The first one is the foliation by circles of the form $\partial D^2 \times \{\mathrm{pt}\}$. The second one is the foliation by circles of the form $\{\mathrm{pt}\} \times S^1$. I assume that by "$S$-transformation" you mean "cut $T_L$ out and glue it back in so that foliations swap palces". Essentially, you are "rotating the gluing map by 90 degrees". Assuming this is correct:
The manifold $\tilde{M}$, obtained by cutting out $T_L$ and gluing it back in with an "$S$-transformation", is homeomorphic to $S^2 \, \tilde{\times}\, S^1$: the $S^2$ bundle over $S^1$ with monodromy the antipodal map.
To see this: After removing $T_L$, what remains is $M^2 \times S^1$. Note that $M^2$ is an interval bundle over the circle. Each interval, crossed with a circle, gives an annulus properly embedded in $M^2 \times S^1$. The $S$-transformation glues a pair of disks to the two boundaries of each such annulus. This is enough to show that $\tilde{M}$ is an $S^2$ bundle over the circle. There are only two of these - the product and the twisted bundle. Since $\tilde{M}$ is not orientable, it is homeomorphic to the twisted bundle.