Is the functor that takes an $R$-algebra to the group of finitely generated projective modules of rank one additive? 
Let $P$ be the covariante functor from the category of commutative $R$-algebras to the category of abelian groups that takes an $R$-algebra $T$ to the group $P(T)$ of the finitely generated projective $T$-modules of rank one.

In Galois Theory and Galois Cohomology of Commutative Rings (CHR), it is said to be a trivial verification that this functor is additive.
I can't find any references to this functor, neither prove it is additive. I can get that if $P(f+g) = P(f) + P(g)$, the inclusion and projection maps build a inverse such that $P(A \oplus B) = P(A) \oplus P(B)$, but I don't know $P(f)$.
Is this related to the Picard Functor? I don't work with schemes or algebraic geometry at all, and found it while looking for references.

Article Definition of Additive Functor: Be $F$ a covariant functor from the category of commutative $R$-algebras to the category of abelian groups. If $J$ is a finite set and $S_j$ is a commutative $R$-algebra for each $j \in J$, the projection maps $p_i: \prod_{j\in J}S_j \rightarrow S_i$ give rise to homomorphisms $F(p_i)$; these in turn give rise to a homomorfism $$\varphi_J: F(\prod_{j \in J}S_j) \rightarrow \prod_{j \in J}F(S_j)$$ defined by $(\varphi_J(x))(i) = F(p_i)(x)$, for $x \in F(\prod_{j \in J}S_j)$ and $i \in J$, viewing an element of a direct product as a function on the index set.

 A: $\newcommand{\Mod}{\mathbf{Mod}}$
Note that by extension of scalars, your maps $p_i: \prod_{j\in J}S_j \to S_i$ also induce a functor $\Mod_{\prod_{j\in J}S_j} \to \prod_{j\in J}\Mod_{S_j}$
It's a classical fact that this functor is an equivalence of categories, its inverse being given by $(M_j)_j \mapsto \bigoplus_j (p_j)^*M_j$
Now this functor is also a symmetric monoidal functor, so it's a symmetric monoidal equivalence.
But now note the following :

Your functor $P(T)$ sends $T$ to $Pic(\Mod_T)$, where $Pic(C)$ is the group of $\otimes$-invertible objects, whenever $(C,\otimes)$ is a symmetric monoidal category.

Indeed, if $P$ is projective of rank one over $T$, then $P\otimes \hom_T(P,T)\cong T$ (you have a natural map $P\otimes_T \hom_T(P,T)\to T$, which is an isomorphism whenever you localize it at a prime ideal, therefore it's an isomorphism), so it's invertible.
Conversely, if $P\otimes_T Q\cong T$, then $\hom_T(P,-)\cong \hom_T(T, Q\otimes -)$, so $P$ is projective; and you can easily check that it's of rank one.
Now it's pretty straightforward to check that for any product of symmetric monoidal categories $\prod_i C_i$, $Pic(\prod_i C_i)\cong \prod_i Pic(C_i)$
From these two facts it follows at once that $P(\prod_j S_j)\cong \prod_j P(S_j)$.
(I think the terminology "additive" is not well-suited for this kind of thing, because as was pointed out, $\mathrm{CAlg}_R$ is not an additive category; one would rather say "product-preserving")
