Ultimately, you make a totally legitimate point; we very well could have defined the term "four-vector" to refer to a type of object that transforms in a different way, but we make the particular definition that we do because it is useful to have a term that refers things that transform like spacetime positions when you change frame. Here are two reasons why:
Fact 1. Given any two four vectors $A^\mu$ and $B^\nu$, the quantity $g_{\mu\nu}A^\mu B^\nu$ is invariant under a change of frame.
Notice that this would not have been true unless $A^\mu$ and $B^\nu$ were four-vectors because the proof of this fact relies on the metric being preserved by Lorentz transformations, and not by other arbitrary things. Here is another reason why the definition is useful
Fact 2. Lots of really useful and physically significant quantities happen to be four-vectors. Take, for example, $J^\mu$ and $A^\mu$ (the current and vector potential) in electromagnetism.
Having said all of this, however, note that there are tons of other quantities that do not transform as four-vectors when one changes frame. In fact, given any representation $\rho$ of the Lorentz group, one often encounters quantities $Q$ that transform as
$$
Q' = \rho(\Lambda) Q
$$
For example, there are objects called Weyl spinors that transform as
$$
\psi' = \rho_\mathrm{weyl}(\Lambda)\psi
$$
when one transforms between frames.
The upshot of all of this is the following
Upshot. Lorentz 4-vectors are not special. However, since every change of reference frame can be associated with a Lorentz transformation, every quantity that you want to transform between frames must necessarily transform in a way the depends, in some way or another, on the Lorentz transformation between the frames. This leads us to not only define four-vectors, but a host of other objects that have specified transformation laws under changes of frame and to give them special names. Doing this is useful because such objects appear all over the place in physics, and we can prove useful properties about objects with certain transformation behaviors.
It is actually possible, and not too difficult, to prove this without expanding the exponentials to first order only.
What you are trying to prove is $S^\dagger \gamma^0 = \gamma^0 S^{-1}$, this is equivalent to
$$ \gamma^0 S^\dagger \gamma^0 = S^{-1} $$
because $( \gamma^0 )^2 = 1$.
Expand $S^\dagger = \sum_n \frac{1}{n!} \left( \frac i 4 \omega_{\mu\nu} \sigma^{\mu\nu\dagger} \right)^n$ and you see that it is sufficient to prove
$$ \gamma^0 \left( \sigma^{\mu\nu\dagger} \right)^n \gamma^0 = \left( \sigma^{\mu\nu} \right)^n \;. $$
Now, the left hand side is equal to $\left( \gamma^0 \sigma^{\mu\nu\dagger} \gamma^0 \right)^n$, hence we only need to prove that
$$ \gamma^0 ( \sigma^{\mu\nu} )^\dagger \gamma^0 = \sigma^{\mu\nu} \;. $$
This is obvious from your equation (2).
Best Answer
$\color{blue}{\textbf{ANSWER A}}\:$ (based on charge invariance, paragraph extracted from Landau)
The answer is given in ACuriousMind's comment as pointed out also by WetSavannaAnimal aka Rod Vance. Simply I give the details copying from "The Classical Theory of Fields", L.D.Landau and E.M.Lifshitz, Fourth Revised English Edition :
(1) Note by Frobenius : We have \begin{equation} dVd(ct)=dx^{1}dx^{2}dx^{3}dx^{4} \tag{01} \end{equation} Now, for the relation between the infinitesimal 4-volumes in Minkowski space \begin{equation} dx'^{1}dx'^{2}dx'^{3}dx'^{4} =\begin{vmatrix} \dfrac{\partial x'_{1}}{\partial x_{1}}& \dfrac{\partial x'_{1}}{\partial x_{2}}&\dfrac{\partial x'_{1}}{\partial x_{3}}&\dfrac{\partial x'_{1}}{\partial x_{4}}\\ \dfrac{\partial x'_{2}}{\partial x_{1}}& \dfrac{\partial x'_{2}}{\partial x_{2}}&\dfrac{\partial x'_{2}}{\partial x_{3}}&\dfrac{\partial x'_{2}}{\partial x_{4}}\\ \dfrac{\partial x'_{3}}{\partial x_{1}}& \dfrac{\partial x'_{3}}{\partial x_{2}}&\dfrac{\partial x'_{3}}{\partial x_{3}}&\dfrac{\partial x'_{3}}{\partial x_{4}}\\ \dfrac{\partial x'_{4}}{\partial x_{1}}& \dfrac{\partial x'_{4}}{\partial x_{2}}&\dfrac{\partial x'_{4}}{\partial x_{3}}&\dfrac{\partial x'_{4}}{\partial x_{4}} \end{vmatrix} dx^{1}dx^{2}dx^{3}dx^{4}=\left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\right\vert dx^{1}dx^{2}dx^{3}dx^{4} \tag{02} \end{equation} where $\:\left\vert\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)/\partial\left(x^{1},x^{2},x^{3},x^{4}\right)\right\vert\:$ the Jacobian, that is determinant of the Jacobi matrix. But the Jacobi matrix is the Lorentz matrix $\:\Lambda\:$ with $\:\det(\Lambda)=+1$, that is \begin{equation} \left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\right\vert=\det(\Lambda)=+1 \tag{03} \end{equation} so \begin{equation} dx'^{1}dx'^{2}dx'^{3}dx'^{4} =dx^{1}dx^{2}dx^{3}dx^{4}=\text{scalar invariant} \tag{04} \end{equation}