I want to show this by using Alexander's Theorem's proof method. So here's what I thought.
As I surger $T$, I have 2 $S^2$. So one bounds $S^1$ and the other bounds $D^2$. By reversing the surgery, I have that $T$ bounds a solid torus $S^1\times D^2$ in $S^3$.
Is it correct?
On
"There's also a proof of this that involves basically following along with what Hatcher did for spheres and... doing it for tori. I don't remember the details, but that's how I did this argument when I looked at the book first."
Let $T$ be an embedded torus in $\mathbb{R}^3$. Then isotope $T$ so that $h:T\to \mathbb{R}$, the height function, is a Morse function. Nothing fancy here, push $T$ around so that none of the saddle points, maximums, and minimums are at the same height. Then slice $T$ along horizontal planes at heights which trap each critical value of $h$ between two planes. Since $T$ is closed and $2$ dimensional, the intersections of $T$ with the horizontal planes will be circles. By cutting $T$ along innermost circles in the intersection with the horizontal planes, pushing the circles off of the planes, and capping the wounds with disks, we surger $T$ into a bunch of components which are isotopic to spheres.
If all of our surgery loops, pre-surgery, bound disks in $T$, then it turns out $T$ is a sphere. This can be seen by induction: The base case is after all of the surgeries. Our manifold is a collection of spheres. The inductive step is a single reverse surgery: by hypothesis will connect a two spheres together, which is always a sphere.
Thus there is a loop $\gamma$ which doesn't bound a disk in $T$ that does bound a disk in $\mathbb{R}^3$. This is the loop OP wanted.
Remark: This is the general proof that no surface $S\subset S^3$ is incompressible
This answer carries through the details of Lee Mosher's comment.
Your goal is to find some loop in the torus that you can do surgery on and get an embedded $S^2$. Equivalently, there is a loop in the torus that bounds an embedded disc in $S^3$ that doesn't interset the torus. (These are equivalent because you can thicken this embedded disc and take its boundary to obtain the $S^1 \times I$ you'll swap out for an $S^0 \times D^2$.). This embedded torus splits the manifold into two sides; call them $M_1$ and $M_2$.
Inspired by the loop theorem, we want to show that there is an embedded loop in $T^2$ such that on one side or the other, it's null-homotopic. Clearly it's null-homotopic in $S^3$. Pick a disc representing that null-homotopy and require that it's transverse to the torus $T^2$ except along its boundary. (You can do this by standard transversality arguments.) Call this map $i$. Pick an innermost circle of $i^{-1}(T^2)$; then because it's innermost, if we restrict $i$ to this circle and the disc it bounds, it now maps only to one of $M_1$ and $M_2$. Suppose for convenience it's $M_1$.
Now that we've got such a loop, we can use the loop theorem to turn the disc it bounds into an embedded one. Unfortunately, I don't really know how to get around using the loop theorem if we're going to do this argument! I could see one trying to prove it for 3-manifolds that live inside $S^3$, but I really don't see how that would help us. But with the loop theorem, we can surger that loop, like you suggest, and invoke Alexander's theorem.
There's also a proof of this that involves basically following along with what Hatcher did for spheres and... doing it for tori. I don't remember the details, but that's how I did this argument when I looked at the book first.