Let $(A, I)$ be a pair. Consider the category $\mathcal{C}$ consisting of \'etale ring maps $A \to B$ such that $A/I \to B/IB$ is an isomorphism. We will show that the category $\mathcal{C}$ is directed and that $A^h = \colim_{B \in \mathcal{C}} B$ with ideal $I^h = IA^h$ gives the desired adjoint. \medskip\noindent We first prove that $\mathcal{C}$ is directed (Categories, Definition \ref{categories-definition-directed}). It is nonempty because $\text{id} : A \to A$ is an object. If $B$ and $B'$ are two objects of $\mathcal{C}$, then $B'' = B \otimes_A B'$ is an object of $\mathcal{C}$ (use Algebra, Lemma \ref{algebra-lemma-etale}) and there are morphisms $B \to B''$ and $B' \to B''$. Suppose that $f, g : B \to B'$ are two maps between objects of $\mathcal{C}$. Then a coequalizer is $$ (B' \otimes_{f, B, g} B') \otimes_{(B' \otimes_A B')} B' $$ which is \'etale over $A$ by Algebra, Lemmas \ref{algebra-lemma-etale} and \ref{algebra-lemma-map-between-etale}. Thus the category $\mathcal{C}$ is directed. \medskip\noindent Since $B/IB = A/I$ for all objects $B$ of $\mathcal{C}$ we see that $A^h/I^h = A^h/IA^h = \colim B/IB = \colim A/I = A/I$. \medskip\noindent Next, we show that $A^h = \colim_{B \in \mathcal{C}} B$ with $I^h = IA^h$ is a henselian pair. To do this we will verify condition (2) of Lemma \ref{lemma-characterize-henselian-pair}. Namely, suppose given an \'etale ring map $A^h \to A'$ and $A^h$-algebra map $\sigma : A' \to A^h/I^h$. Then there exists a $B \in \mathcal{C}$ and an \'etale ring map $B \to B'$ such that $A' = B' \otimes_B A^h$. See Algebra, Lemma \ref{algebra-lemma-etale}. Since $A^h/I^h = A/IB$, the map $\sigma$ induces an $A$-algebra map $s : B' \to A/I$. Then $B'/IB' = A/I \times C$ as $A/I$-algebra, where $C$ is the kernel of the map $B'/IB' \to A/I$ induced by $s$. Let $g \in B'$ map to $(1, 0) \in A/I \times C$. Then $B \to B'_g$ is \'etale and $A/I \to B'_g/IB'_g$ is an isomorphism, i.e., $B'_g$ is an object of $\mathcal{C}$. Thus we obtain a canonical map $B'_g \to A^h$ such that $$ \vcenter{ \xymatrix{ B'_g \ar[r] & A^h \\ B \ar[u] \ar[ur] } } \quad\text{and}\quad \vcenter{ \xymatrix{ B' \ar[r] \ar[rrd]_s & B'_g \ar[r] & A^h \ar[d] \\ & & A/I } } $$ commute. This induces a map $A' = B' \otimes_B A^h \to A^h$ compatible with $\sigma$ as desired. \medskip\noindent Let $(A, I) \to (A', I')$ be a morphism of pairs with $(A', I')$ henselian. We will show there is a unique factorization $A \to A^h \to A'$ which will finish the proof. Namely, for each $A \to B$ in $\mathcal{C}$ the ring map $A' \to B' = A' \otimes_A B$ is \'etale and induces an isomorphism $A'/I' \to B'/I'B'$. Hence there is a section $\sigma_B : B' \to A'$ by Lemma \ref{lemma-characterize-henselian-pair}. Given a morphism $B_1 \to B_2$ in $\mathcal{C}$ we claim the diagram $$ \xymatrix{ B'_1 \ar[rr] \ar[rd]_{\sigma_{B_1}} & & B'_2 \ar[ld]^{\sigma_{B_2}} \\ & A' } $$ commutes. This follows once we prove that for every $B$ in $\mathcal{C}$ the section $\sigma_B$ is the unique $A'$-algebra map $B' \to A'$. We have $B' \otimes_{A'} B' = B' \times R$ for some ring $R$, see Algebra, Lemma \ref{algebra-lemma-diagonal-unramified}. In our case $R/I'R = 0$ as $B'/I'B' = A'/I'$. Thus given two $A'$-algebra maps $\sigma_B, \sigma_B' : B' \to A'$ then $e = (\sigma_B \otimes \sigma_B')(0, 1) \in A'$ is an idempotent contained in $I'$. We conclude that $e = 0$ by Lemma \ref{lemma-idempotents-determined-modulo-radical}. Hence $\sigma_B = \sigma_B'$ as desired. Using the commutativity we obtain $$ A^h = \colim_{B \in \mathcal{C}} B \to \colim_{B \in \mathcal{C}} A' \otimes_A B \xrightarrow{\colim \sigma_B} A' $$ as desired. The uniqueness of the maps $\sigma_B$ also guarantees that this map is unique. Hence $(A, I) \mapsto (A^h, I^h)$ is the desired adjoint.