Why is $F^*(\mathcal L) = \mathcal L^{\otimes p}$ where $F$ is the absolute Frobenius and $\mathcal L$ is an invertible sheaf?

It is true that $F^{-1}O_S=O_S$, but in the tensor product for $F^*L$, the $O_S$-module structure of the right factor $O_S$ is not given by the identity $O_S\to O_S$, it is the Frobenius $F^{\#}$. It would be less confusing to write $F: T \to S$, and consider the tensor product $F^{-1}L\otimes_{F^{-1}O_S} O_T$.

Further remark on the construction of $F^*L$: you should not use the very definition, but the property that if $S=\mathrm{Spec}(A)$, $T=\mathrm{Spec}(B)$ and $L$ is any quasi-coherent sheaf on $S$ associated to an $A$-module $M$, then $F^*L$ is just the quasi-coherent sheaf on $T$ associated to the $B$-module $M\otimes_A B$.

A proof of $F^*L\simeq L^{\otimes p}$ is as you suggested: consider (the isomorphism class of) the sheaf $L$ as an element of $\mathrm{Pic}(S)=H^1(S, O_S^*)$. Then the effect of taking $F^*$ is given by $F^{\#}: O_S^* \to O_S^*$.

Anothe proof gives directly an isomorphism as follows. Let $A$ be an $\mathbb F_p$-algebra. Denote by $\rho=F^{\#}$ the absolute Frobenius on $A$, and by $A_\rho$ the $A$-algebra whose ring is $A$, but the $A$-algebra structure is given by $\rho : A\to A_\rho$. Similarly, for any $A$-module $M$, let $M_\rho$ be $M$ endowed with the structure of $A$-module via $a*x=\rho(a)x$. This is also an $A_\rho$-module whose structure is exactly that of $M$ as $A$-module. Let $S(M_\rho)$ be the symetric algebra over $A_\rho$ associated to $M_\rho$. We have an $A$-bilinear map $$ \phi: M \times A_\rho \to S(M_\rho), \quad (x, b)\mapsto bx^p.$$ Actually $\phi(ax, b)=b(ax)^p=ba^px^p=\rho(a)\phi(x,b)=a*\phi(x,b)$, and $\phi(x, a*b)=\phi(x, a^pb)=a^p\phi(x,b)=a*\phi(x,b)$. So $\phi$ induces a $A$-linear map $$ M\otimes_A A_\rho \to S(M_\rho), \quad x\otimes b\mapsto bx^p.$$ This map is $A_\rho$-linear too. When $M$ is free of rank one, this induces an isomorphism onto the degree $p$ component of the symetric algebra which is also $M^{\otimes p}$. Continued in the comments.

Even more simple, and more general:

Let $S$ be a ringed space of $\mathbb{F}_p$-algebras, and let $F : S \to S$ be the absolute Frobenius. Then for every line bundle $\mathcal{L}$ on $S$ we have $F^* \mathcal{L} \cong \mathcal{L}^{\otimes p}$.

Proof. Since $F^*$ is left adjoint to $F_*$ (which could be seen as a definition), for every sheaf of modules $M$ on $S$ we have to prove $\hom(\mathcal{L},F_* M) \cong \hom(\mathcal{L}^{\otimes p},M)$, naturally in $M$. For $\mathcal{L}=\mathcal{O}_S$ this is trivial, both sides identify with $\Gamma(M)$, and in general $\mathcal{L}$ looks locally like this. So we only have to observe that the isomorphisms glue, i.e. that for every isomorphism $\gamma : \mathcal{O}_S \to \mathcal{O}_S$ resp. unit $\gamma \in \Gamma(\mathcal{O}_S)^*$ the diagram

$$\begin{matrix} \hom(\mathcal{O}_S,F_* M) & \cong & \Gamma(F_* M) & = & \Gamma(M) & \cong & \hom(\mathcal{O}_S^{\otimes p},M) \\ \gamma^* \downarrow ~ & & ~ \downarrow \cdot \gamma ~ & & ~ \downarrow \cdot \gamma^p & & ~ \downarrow (\gamma^{\otimes p})^* \\\hom(\mathcal{O}_S,F_* M) & \cong & \Gamma(F_* M) & = & \Gamma(M) & \cong & \hom(\mathcal{O}_S^{\otimes p},M) \end{matrix}$$

commutes, which can be checked directly by the definition of $F$.