For a commutative ring $R$, consider the formal group $\widehat{\mathbb{G}}\_m$ over $R$ that is the completion of $\mathbb{G}\_{m, R}$ along its identity section (naively, $\widehat{\mathbb{G}}\_m$ is the formal group law $F(X, Y) = XY + X + Y \in R[[X, Y]]$). Now let $R$ be Noetherian and such that there is a nilpotent ideal $I \subset R$ with $I^n = 0$ such that $R/I = \mathbb{Z}$. Suppose one has a smooth commutative $R$-group scheme $G$ (of finite type if that's relevant) equipped with an isomorphism $G\_{R/I} \cong \mathbb{G}\_m$ and that one considers the formal completion $\widehat{G}$ of $G$ along its identity section. Is $\widehat{G}$ necessarily isomorphic to $\widehat{\mathbb{G}}\_m$ as formal groups over $R$?



Infinitesimal deformations of the formal group of $\mathbb{G}_m$
Are you trying to justify some of the assertions about the Tate curve in Katz-Mazur? (Would help to know the motivation for the question.) Anyway, the affirmative answer is part of the "standard" infinitesimal deformation theory for tori as developed in SGA3. In fact, the intervention of formal groups is a red herring, as is the hypothesis $R/I = \mathbf{Z}$ (the latter being what suggests to me that you are really trying to understand something about the Tate curve): $G$ is *uniquely* $R$-isomorphic to $\mathbf{G}\_m$ in a manner lifting the given such isomorphism over $R/I$. 


Indeed, clearly $G$ is affine, as $G\_0 := G \bmod I$ is affine and $I^n=0$, so we can apply the deformation theory of tori to the given isomorphism $f\_0: \mathbf{G}\_m \simeq G\_0$. By deformation theory considerations with split relative tori and Hochschild cohomology (in the recently-published proceedings of the Luminy summer school on SGA3, see either item 3(c) in Chapter IV of Oesterle's article on group schemes of multiplicative type, or see Corollary B.2.6 and the comment immediately after its statement in the article on reductive group schemes), this $f\_0$ lifts uniquely to an $R$-homomorphism $f:\mathbf{G}\_m \rightarrow G$, and $f$ is an isomorphism by the fibral isomorphism criterion (with source and target fppf over the base) since $f\_0$ is an isomorphism. 


