I think you pretty much have the solution. You've just written the pieces in a sequence that may be a bit incompletely explained.
Your sequence of equations/inequalities is equivalent to
the following set of statements:
\det(A) &= \det(T_\alpha) \cdot \det(A) \\
\det(T_\alpha) \cdot \det(A) &= \det(T_\alpha A) \\
\det(T_\alpha A) &> 0.
You clearly know how to explain why the two equations are true.
If you can just show that the last inequality is true,
then you can put it all together to prove that if
$\det(T_\alpha A) > 0$ then $\det(A) > 0.$
So consider what you know about the columns of $T_\alpha A$
and see what that tells you about its determinant,
and you can show that if a rotation $T_\alpha A$ exists as described
in the problem, then $\det(A) > 0.$
(Perhaps you've already done this but thought it was too obvious to state;
I would say it's reasonably obvious, but not so obvious that it's not
For the other direction, the sequence in which you wrote the
equations and inequalities is already a good one for explanation.
You just need to establish somehow that if $\det(A) > 0$
then $v_1 \neq 0$ and there is some rotation $T$ that will rotate
$v_1$ to a vector in the same direction as $e_1,$
that is, $Tv_1 = ce_1$ for $c\in \mathbb R$ and $c>0.$
Call that rotation $T_\alpha,$ and from your formulas you can
show that if $\det(A) > 0$ then $\det(T_\alpha A) > 0.$
Now look at how you compute $\det(T_\alpha A)$ and what you know
about the coordinates of $T_\alpha v_1,$
and see what that tells you about the second coordinate of $T_\alpha v_2$
(recalling that all you need to show now is that $T_\alpha v_2$
points somewhere in the upper half-plane).
And you're still right, it's pretty much obvious, the only trick is to convince the reader that you haven't just forgotten about one of the fact that makes it obvious.