Group homomorphisms out of the additive group is formal completion at the unipotent cone

Let $G$ be an affine algebraic group over an algebraically closed field of characteristic zero $k$. Then, the functor $\text{Hom}_{grp}(\mathbb{G}_a, G)$ is representable by a colimit of schemes (this is in Conrad/Gabber/Prasad, Pseudo-reductive groups A.8.13). Intuitively, this colimit should be something like the infinitesimal neighborhood of the unipotent cone.

Is an argument for this latter intuitive description written up anywhere? I have a partial argument as follows. Namely, consider the map $\text{Hom}_{grp}(\mathbb{G}_a, G) \rightarrow \text{Hom}_{grp}(\mathbb{Z}, G) = G$. First, if we just look at $k$-points, the classical theory (i.e. by looking at eigenvalues on representations) tells us that the image is the unipotent cone $U$. We claim that this map actually defines a subfunctor, and it is the formal completion along the unipotent cone.

Here's where I get kind of stuck. I want to just say, take some let $R$ be a ring, and $R/I$ its reduced ring, and on $R$-points, take some $u \in G(R)$ whose reduction in $G(R/I)$ is unipotent i.e. it is in $U(R/I)$. Then, fixing some faithful representation $V$, there $1 - u \in \text{End}_k(V) \otimes_k R$ is nilpotent, so $\log$ makes sense, and we can define a homomorphism $\mathbb{G}_a(S) \rightarrow G(S)$ via the exponential map.

Note, I know there's some issue about sheafy Hom here I glossed over, but I claim it's not a problem and I don't want to clutter the exposition.

The problem is I'm not sure how to show this is a subfunctor. I want to claim that the exponential map is somehow unique, and probably, this is possible by unwinding the Hopf algebra comultiplications, but I cant think of a smart way to do this without the problem quickly becoming unwieldy.

So, two questions: (1) Does this partial argument sound reasonable so far? (2) Can this argument be completed, or is it somewhere in the literature already?