Let $X$ be a scheme (probably a locally ringed space would work just as well), $\mathcal{L}$ be a line bundle (i.e. locally free sheaf of rank one or invertible sheaf) on $X$, and $s \in \Gamma(X,\mathcal{L})$ be global section.
Let me define a few things before I ask my question :
We say that the $s$ is regular if the map $$ \Phi : \mathcal{O}_X \to \mathcal{L} \;\;\;\;\;\; f \mapsto fs $$ is injective.
Associated to $s$ there is a natural evaluation map : $$ \Psi : \mathcal{L}^\vee \to \mathcal{O}_X \;\;\;\; u \mapsto u(s) $$ The image of $\Psi$ is an $\mathcal O_X$-ideal. $$ I(s) = Im(\Psi) $$
How does one prove that $I(s)$ is an invertible sheaf if and only if $s$ is regular ?
Here is my tentative to implement Mohan's answer in detail :
Let $(U_i)_{i\in I}$ be a covering of $X$ by affine schemes (say $U_i = Spec(A_i)$) which trivialise $\mathcal{L}$. This means that $\mathcal{L}_{|U_i} \cong \widetilde{L_i}$ where $L_i = e_iA_i$ is a free $A_i$-module of rank $1$.
We put : $s_i = s_{|U_i} \in \mathcal L(U_i) = L_i$, we can write $s_i = t_i e_i$ with $t_i \in A_i$.
By the equivalence of categories between $A_i$-modules and $\mathcal{O}_{U_i}$-modules, $\Psi_{|U_i}$ comes from the map $\Psi_i : L_i^\vee \to A_i$ defined by $u \mapsto u(s_i) = t_iu(e_i)$.
Since the map $L_i^\vee \to A_i$, $u \mapsto u(e_i)$ is an isomorphism we see that $Im(\Psi_i) = t_iA$ and $t_iA$ is free of rank $1$ if and only if $t_i$ is not a zero divisor in $A_i$.
We get that $I(s)_{|U_i} = \widetilde{t_iA_i}$ then $I(s)$ being locally free of rank $1$ is equivalent to $I(s)_{|U_i}$ begin locally free of rank $1$ for each $i \in I$ which in turn is equivalent to each $t_iA_i$ being locally free of rank $1$ which is equivalent to each $t_i$ begin a non-zero divisor.
Now we recall that a map of sheaves is injective if and only if it is so locally. So $\Phi$ is injective if and only if $\Phi_{|U_i}$ is injective for all $i \in I$. But $\Phi_{|U_i}$ comes from the map $\Phi_i : A_i \to L_i= e_iA$ given by $a \mapsto as_i = at_ie_i$. and this map is injective if and only if $t_i$ is not a zero divisor.
This proves the result.
Invertibility is a local statement. Locally your line bundle is isomorphic to the ring of functions, say $A$. Then you have the map $A\to A$ given by multiplication by an element $s\in A$. Regular means, this element is not a zero divisor in $A$ and thus dually you get the the map, which looks exactly the same and the image is generated by $s$. But $s$ is a non-zero divisor means $sA$ is isomorphic to $A$.