Funny problem. There is a number of ways it can be treated, starting from elementary (working with real & imaginary parts) & proceeding to more advanced (including some more geometric approaches).
Your questions 1 & 2 read pretty much the same to me, & the only sensible answer I can provide is 'because working w/ z & $\bar{z}$ concurrently allows us to use Linear Algebra' - without getting messy, at least, which is something that the solution using real & imaginary parts cannot claim. That aside, it's clearly a trick (but then again, what isn't?).
I understand your question about the determinant condition working out great in $\mathbb{C}^2$ but possibly being an overkill here, & it's a valid question. I propose to explain why the condition is in fact both sufficient & necessary using subspaces. I'll warn you that this boils down to exactly working w/ real & imaginary parts, albeit after having recast your problem in the $2\times2$ matrix form you give. From a certain POV, that's keeping the worst of both worlds: passing to that form is unintuitive & working with real/imaginary parts is cumbersome. Nevertheless, I find it far more intuitive than either solution, for reasons that might become clear once I stop running my mouth & do some math.
Let's consider any two complex numbers $z = z_R + {\rm i} z_I$ & $z' = z'_R + {\rm i} z'_I$. I embed (z,z') into $\mathbb{R}^4$ by writing $(z_R,z_I,z'_R,z'_I)$ for $(z,z')$ - this is a homomorphism between $\mathbb{C}^2$ and $\mathbb{R}^4$. Then, the pair $(z,\bar{z})$ becomes $(z_R,z_I,z_R,-z_I) = z_R(1,0,1,0) + z_I(0,1,0,-1) = z_R {\rm e_1} + z_I {\rm e_i}$, with ${\rm e_{1,i}}$ the obvious vectors in $\mathbb{R}^4$. Hence, all possible pairs $(z,\bar{z})$ constitute a $2-$D subspace $\mathbb{E}$ of $\mathbb{R}^4$, spanned by ${\rm e_{1,i}}$.
Write, now, $T : \mathbb{C}^2 \to \mathbb{C}^2$ for the linear operator represented by the matrix $\left(\begin{array}{cc}a&b\\\bar{b}&\bar{a}\end{array}\right)$. Then, the restriction $\left.T\,\right\vert_\mathbb{E} : \mathbb{E} \to \mathbb{E}$ is well defined, that is, $T$'s range falls indeed within $\mathbb{E}$. Indeed, we have (easy calculation)
$$
T {\rm e_1} = (a_R+b_R){\rm e_1} + (a_I+b_I){\rm e_i} ,
\\
T {\rm e_i} = (b_I-a_I){\rm e_1} + (a_R-b_R){\rm e_i} .
$$
Equipping $\mathbb{E}$ with the basis ${\rm e_{1,i}}$ and expressing $T$ in that basis, then, we obtain the $2\times2$ matrix
$$
\left(\begin{array}{cc}
a_R+b_R & b_I-a_I
\\
a_I+b_I & a_R-b_R
\end{array}\right) ,
$$
which is precisely what you'd obtain had you chosen to recast the original equation in terms of real & imaginary parts. (The reason is obvious, I believe.) Its determinant is, naturally, $|a|^2-|b|^2$.
Note, finally, that the equation for $(z,\bar{z})$ you report has a right-hand side that also falls in $\mathbb{E}$ (also for obvious intuitive reasons). Hence, the equation is solvable as long as $\left.T\,\right\vert_\mathbb{E} : \mathbb{E} \to \mathbb{E}$ is a bijection, i.e., as long as the aforementioned determinant is nonzero. & that's why the determinant condition is also necessary, not just sufficient.