Below I will bring a passage from Heat Kernels by Wolfgang Arendt (Theorem 4.3.3, page 52). I need to understand it and write a more verbose report based on the chapter, however I am stuck at this part. I managed to repeat most of the steps, but the "this shows" (highlighted with !!!) sentence confuses me. How was the author able to do this argument? What does the observation do in this proof? I see that it has something to do with dual spaces and Banach adjoints ($\langle Ax,x^*\rangle = \langle x,A^*x^*\rangle$) and that the integral should be understood as a $L^1(\Omega)^* \cap L^\infty(\Omega)^*$ function (aka $\langle x, x^*\rangle$), but I can't understand where all this is going.
Let $S \in \mathcal{L}(L^1(\Omega))$ be positive such that $\lVert S\rVert_{\mathcal{L}(L^1(\Omega))}, \lVert S\rVert_{\mathcal{L}(L^\infty(\Omega))} \leq M$.
We have to show that there exists a unique operator $T \in \mathcal{L}(L^\infty(\Omega))$ such that $T^*f = Sf$ for all $f \in L^1(\Omega) \cap L^\infty(\Omega)$. Then we let $S_\infty = T^*$. Uniqueness of T is easy to see. In order to prove its existence we note that the hypothesis implies that $S(L^1(\Omega) \cap L^\infty(\Omega)) \in S(L^1(\Omega) \cap L^\infty(\Omega))$ and $\lVert Sf\rVert_{L^p(\Omega)} \leq M\lVert f\rVert_{L^p(\Omega)}$ for $p = 1, \infty$ and for all $f \in L^1(\Omega) \cap L^\infty(\Omega)$.
Observe that a function $g \in L^\infty(\Omega)$ is in $L^1(\Omega) \cap L^\infty(\Omega)$ and $\lVert g \rVert_{L^1(\Omega)} \leq M$ if an only if
$\lvert \int_\Omega g(x)\phi(x)dx \rvert \leq M\lVert \phi \rVert_{L^\infty(\Omega)}$
for all $\phi \in L^1(\Omega) \cap L^\infty(\Omega)$. (!!!) This shows that the adjoint $S^* \in \mathcal{L}(L^\infty(\Omega))$ of $S$ leaves $L^1(\Omega) \cap L^\infty(\Omega)$ invariant and that $\lVert S^*f\rVert_{L^1(\Omega)} \leq M\lVert f\rVert_{L^1(\Omega)}$ for all $f \in L^1(\Omega) \cap L^\infty(\Omega)$ (!!!). By density of $L^1(\Omega) \cap L^\infty(\Omega)$ in $L^1(\Omega)$, there exists a unique operator $T \in \mathcal{L}(L^1(\Omega))$ such that $Tf = S^*f$ for all $f \in L^1(\Omega) \cap L^\infty(\Omega)$. This implies that $T^*g = Sg$ for all $g \in L^1(\Omega) \cap L^\infty(\Omega)$.
Let us denote by $T$ the anti-linear isomorphism $$T: L^\infty(\Omega)\rightarrow (L^1(\Omega))^*, g\mapsto \left( \varphi\mapsto \int_\Omega \overline{g(x)}\varphi(x)dx\right).$$ When we are talking about $S^*f$ for $f\in L^\infty (\Omega)$, then we really mean $T^{-1} S^* Tf$.
So, the actual claim we want to prove is that $(T^{-1} S^* T)(f)\in L^1(\Omega)$ and $$\Vert (T^{-1} S^* T)(f)\Vert_{L1(\Omega)}\leq M \Vert f \Vert_{L^1(\Omega)}.$$ Now we compute for $\phi\in L^1(\Omega)\cap L^\infty(\Omega)$ \begin{align*} \left\vert \int_\Omega \overline{(T^{-1} S^* Tf)(x)} \phi(x) dx \right\vert &=\vert (S^* T f)(\phi)\vert \\ &=\vert (Tf)(S\phi)\vert \\ &=\left\vert \int_\Omega \overline{f(x)} (S\phi)(x) dx \right\vert \\ &\leq \Vert f \Vert_{L^1(\Omega)} \Vert S\phi\Vert_{L^\infty(\Omega)} \\ &\leq (M \Vert f\Vert_{L^1(\Omega)}) \Vert \phi\Vert_{L^\infty(\Omega)}. \end{align*} Using your fact with $g=\overline{T^{-1} S^* T f}$ yields $$\Vert T^{-1} S^* T f\Vert_{L^1(\Omega)}= \Vert \overline{T^{-1} S^* Tf}\Vert_{L^1(\Omega)}\leq M \Vert f\Vert_{L^1(\Omega)}.$$