**ALERT.** I was alerted to a problem with the following by an e-mail. Unlike the affine case, in the global case, the formal completion of a smooth $S$-scheme along a section may fail to split. Thus, in the examples of $X$ and $Y$ below, the formal completions may be **non-isomorphic**. In fact, likely they are non-isomorphic. So the following example is probably **wrong**.

There cannot be a "global" result in complete generality, but there certainly can be results that are "local" over $S$. Here is one global counterexample. Let $S$ be $\mathbb{P}^1$. Let $Z$ be $S\times S$, and let $\Delta:S \to Z$ be the diagonal morphism. Let $(\pi:X\to Z,\chi:Z\to X)$ be the vector bundle of rank $1$ such that $\chi^*\mathcal{O}_X(\chi(Z))$ is isomorphic to $\text{pr}_1^*\mathcal{O}_S(0)\otimes \text{pr}_2^*\mathcal{O}_S(2)$ as an invertible sheaf on $Z=S\times S$. Similarly, let $(\rho:Y\to Z,\upsilon:Z\to Y)$ be the vector bundle of rank $1$ such that $\upsilon^*\mathcal{O}_Y(\upsilon(Z))$ is isomorphic to $\text{pr}_1^*\mathcal{O}_S(1)\otimes \text{pr}_2^*\mathcal{O}_S(1)$.

Via the projection $\text{pr}_1:Z\to S$, both the $Z$-schemes $X$ and $Y$ are also $S$-schemes. Define $\chi':S\to X$ and $\upsilon':S\to Y$ to be $\chi\circ \Delta$, respectively $\upsilon\circ \Delta$. The point is that every nonempty, effective Cartier divisor in $X$, respectively $Y$, has nonempty intersection with $\chi'(S)$, respectively with $\upsilon'(S)$.

Let $f:U\to X$ be an étale morphism such that $\chi':S\to X$ lifts to $\chi'':S\to U$ with $f\circ \chi'' = \chi'$. There exists a quasi-compact, separated, connected, Zariski open subset of $U$ containing $\chi''(S)$. Thus, without loss of generality, assume that $U$ is quasi-compact, separated and connected. By Grothendieck's formulation of Zariski's Main Theorem (or more explicit arguments), $f$ factors as an open immersion, $i:U\to \overline{U}$, composed with a finite morphism, $\overline{f}:\overline{U}\to X$, where $\overline{U}$ is normal. But then, by purity, the branch divisor of $\overline{f}$ is an effective Cartier divisor. Since $\chi'$ lifts to $\chi''$, $\chi'(S)$ is disjoint from the branch divisor. But then, by the previous paragraph, the branch divisor is empty. Thus $\overline{f}$ is everywhere étale. But, since $X$ has trivial étale fundamental group, this implies that $\overline{f}$ is an isomorphism. Therefore $f:U\to X$ is an open immersion.

The same argument holds for $Y$. So for a pair $(\sigma:U\to S,\tau:S\to U)$, if there exist étale $S$-morphisms, $f_X:U\to X$ and $f_Y:U\to Y$ compatible with $\chi'$, $\upsilon'$ and $\tau$, then, up to shrinking $U$, both $f_X$ and $f_Y$ are open immersions. Moreover, since the complement of $U$ in $X$, respectively $Y$, is disjoint from $\chi'(S)$, respectively $\upsilon'(S)$, the complement has codimension $2$. Thus $X$ and $Y$ are $S$-isomorphic away from codimension $2$.

Now this is a serious problem. An isomorphism away from codimension $2$ induces an isomorphism of Picard groups and of the classes of the dualizing sheaf inside the Picard groups. By construction, the pullback map $\pi^*:\text{Pic}(Z)\to \text{Pic}(X)$, respectively $\rho^*$, is an isomorphism. The isomorphism of Picard groups of $X$ and $Y$ induced from this $S$-isomorphism is compatible with these pullback isomorphisms. Also $\omega_X$ is isomorphic to
$$\pi^* ( \text{pr}_1^*\mathcal{O}_S(-2)\otimes \text{pr}_2^*\mathcal{O}_S(-4)),$$ whereas $\omega_Y$ is isomorphic to
$$\pi^*(\text{pr}_1^*\mathcal{O}_S(-3)\otimes \text{pr}_2^*\mathcal{O}_S(-3)).$$
This contradiction proves that there cannot be a pair $(f_X,f_Y)$ as above.

Finally, I claim that the formal completions of $X$ and $Y$ along the image of $S$ are isomorphic. First of all, the normal sheaf of $S$ in $X$, respectively of $S$ in $Y$, fits into a short exact sequence,
$$ 0 \to \Delta^*N_{\Delta(S)/Z}\to \chi^*N_{\chi(S)/X} \to \chi^*N_{Z/X} \to 0,$$
respectively,
$$ 0 \to \Delta^*N_{\Delta(S)/Z}\to \upsilon^*N_{\upsilon(S)/Y} \to \upsilon^*N_{Z/Y} \to 0.$$
Of course, in both cases this is just,
$$
0 \to \mathcal{O}_S(2) \to \mathcal{O}_S(2)^{\oplus 2} \to \mathcal{O}_S(2) \to 0. $$
So the normal sheaves are isomorphic. But also, the infinitesimal extensions are trivial since there is a retraction induced by the projections to $S$.

**Edit.** I realized that there is an easier example underlying the example above. Again let $S$ be $\mathbb{P}^1$. Now let $X$ be the $S$-scheme $S\times S$ via the first projection, $\text{pr}_1$, and let the section $\chi$ be the diagonal morphism. Next, let $(\rho:Y\to S, \upsilon:S\to Y)$ be the rank $1$ vector bundle over $S$ with $\upsilon^*\mathcal{O}_Y(\upsilon(S))$ isomorphic to $\mathcal{O}_S(2)$, and where $\upsilon$ is the zero section. These $S$-schemes with section are formally isomorphic as above. However, every nonempty, effective Cartier divisor in $S\times S$ intersects the diagonal $\chi(S)$. Thus, as in the above example, there is no $S$-scheme with section $(\sigma:U\to S,\tau:S\to U)$ and pair of étale $S$-morphisms, $f_X:U\to X$, $f_Y:U\to Y$ that is compatible with $\chi$, $\upsilon$, and $\tau$.