Roughly, the trick is not to view $L$ as an operator on $L^2$, but on $C^0$
I will use the following version of Krein-Rutmann which is proven in "Du, Yihong: Order Structure and Topological Methods in Nonlinear Partial Differential Equations, Vol. 1: Maximum Principles and Applications.":
Let $X$ be a Banach space, $C \subset X$ a solid cone (i.e. a cone with nonempty interior) and $T : X \longrightarrow X$ a compact linear operator which is strongly positive, i.e. $Tu \in C$ if $u \in C$. Then the spectral radius $r(T)$ fulfills $r(T) > 0$ and is a simple eigenvalue admitting an eigenvector $\psi \in \mathrm{int} C$ and there is no other eigenvalue that admits an eigenvector in $C$. Furthermore, all other eigenvalues $\lambda$ fulfill $|\lambda| < r(T)$.
By partial integration, you easily show that the operator is bounded from below, so $L + \alpha$ is strongly positive from $C^2$ to $C^0$ for some $\alpha$ big enough. It is also well-known that the inverse $T := (L + \alpha)^{-1}$ exists and is a compact operator on $C^0$. By the strong maximum principle, $Lu>0$ implies $u>0$, so $T$ is a strongly positive operator.
For the strong maximum principle, you can consult Evans: Partial Differential Equations, for example. To get the statement on a manifold instead of an area in $\mathbb{R}^n$, use a partition of unity.
Now it is easy to show that the set of positive functions is actually a solid cone in $C^0$ (even though it has empty interior in $L^2$), so we can apply the Krein-Rutmann theorem.
By elliptic regularity, every $L^2$ eigenfunction is $C^\infty$ and because the manifold is compact, is bounded, hence in $C^0$. Conversely, every $C^0$ Eigenfunction is in $L^2$, again because the domain is bounded. Hence the eigenvectors and eigenfunctions of $L$ are the same, whether viewed as operator on $L^2$ or on $C^0$