Let $A$ be a ring, and let $S$ be the scheme $\textrm{Spec }A$. Then the $S$-scheme $G = \textrm{Spec }A[T]$ has the structure of a group scheme over $S$, via the morphisms of $S$-schemes $$m: G \times_S G \rightarrow G, \iota: G \rightarrow G, e: S \rightarrow G$$
whose corresponding $A$-algebra homomorphisms are given respectively by
$$m^{\ast}: A[T] \rightarrow A[X,Y], \space \space f(T) \mapsto f(X+Y)$$ $$\iota^{\ast}: A[T] \rightarrow A[T], \space \space f(T) \mapsto f(-T)$$
$$e^{\ast}: A[T] \rightarrow A, \space \space f(T) \mapsto f(0)$$
If $X$ is any scheme over $S$, then we can identify $$G(X) = \textrm{Hom}_{S-\textrm{sch}}(X,G) = \textrm{Hom}_{A-\textrm{alg}}(A[T], \mathcal O_X(X)) = \mathcal O_X(X)$$
and the morphisms endow $G(X) = \mathcal O_X(X)$ with a group structure identical to the existing additive group structure on $\mathcal O_X(X)$.
Now, what if we replace the affine scheme $S$ with an arbitrary scheme? Does there exist an $S$-group scheme $G$ such that $G(X) = (\mathcal O_X(X),+)$ for all $S$-schemes $X$? The first thing I thought to try was $\textrm{Spec } \mathcal O_S(S)[T]$, but this probably isn't the correct object.