We need two lemmas:
$$B\mathrm e^A B^{-1}=\mathrm e^{BAB^{-1}}\tag{1}$$ (see here for the proof).
And the fact that the gamma matrices satisfy $$(\gamma^{\mu})^\dagger=\gamma^0\gamma^\mu\gamma^0\tag{2}$$ (the proof is easy; one just considers the cases $\mu=0$ and $\mu=i$ separately).
Here, $S$ is defined as $$ S=\exp\left[-\frac i2 \omega_{\mu\nu}\gamma^{[\mu}\gamma^{\nu]}\right] $$ where $\gamma^\mu$ are the Dirac gamma matrices.
Using $(1)$ together with $(\gamma^0)^2=1$, we can see that $$ \gamma^0 S\gamma^0\overset{(1)}=\exp\left[-\frac i2 \omega_{\mu\nu}\gamma^0\gamma^{[\mu}\gamma^{\nu]}\gamma^0\right] \tag{A} $$
Now, we use $(2)$ together with $(\gamma^0)^2=1$ to conclude that $$ \gamma^0\gamma^{[\mu}\gamma^{\nu]}\gamma^0=\gamma^0\gamma^{[\mu}\gamma^0\gamma^0\gamma^{\nu]}\gamma^0\overset{(2)}=-(\gamma^{[\mu}\gamma^{\nu]})^\dagger\tag{B} $$
Finally, using $(\mathrm A)$ and $(\mathrm B)$, we see that $$ \gamma^0 S\gamma^0=\exp\left[+\frac i2 \omega_{\mu\nu}(\gamma^{[\mu}\gamma^{\nu]})^\dagger\right]=S^\dagger $$ as we wanted to prove.