Let $(A, \wedge, \vee)$ be a distributive lattice and $I \subsetneq A$ a maximal ideal. Then $I$ is prime, that is, $a \wedge b \in I \implies (a \in I \text{ or } b \in I)$.
Suppose that $a \wedge b \in I$, $a \notin I$. Then the ideal $\overline{I}$ generated by the set $I \cup \{a\}$ must be equal to $A$, since it includes $I$ and $I$ is maximal. From there, the proof I've found on planetmath infers that there must be some $p \in I$ such that $b \leq p \vee a$. I can see how $b \in I$ follows, but I don't see the implication from $\overline{I} = A$ to $\exists p \in I : b \leq p \vee a$.
To get a representation of $a \vee b$ as $p \wedge a$ for some $p \in I$ (which would solve the issue, since $a \vee b \in \overline{I}$),, I've tried representing $\overline{I}$ explicitly, for example as $\overline{I} = I \cup \{p \vee a \}_{p \in I \cup \{a\}} \cup \{q \wedge a \}_{q \in A}$. But those kinds of sets turn out to not necessarily be ideals (i.e. in the example set, it turns out $(p \vee a)\wedge q$ with $p \in I, q \in A$ might be excluded from it - including it would necessitate adding $\{p \wedge q \}_{q \in A}^{p \in \overline{I}}$, but it seems to me that would just generate more problems).
The ideal generated by a subset $S$ of a lattice $A$ can be described as $J=\{x\in A: x\leq s_1\vee\dots\vee s_n \text{ for some }x_1,\dots,x_n\in S\}$. Indeed, it is clear that any ideal containing $S$ must contain $J$, and conversely $J$ is an ideal since if $x\leq s_1\vee\dots\vee s_n$ and $y\leq s_1'\vee\dots \vee s_m'$ then $x\vee y\leq s_1\vee\dots \vee s_n\vee s_1'\vee\dots\vee s_m'$.
So in your setup, if $\overline{I}=A$, then in particular $b\in \overline{I}$, so there exist $s_1,\dots,s_n\in I\cup\{a\}$ such that $b\leq s_1\vee\dots\vee s_n$. Now let $p$ be the join of all the $s_i$ that are in $I$. Since $I$ is an ideal, $p\in I$, and $s_1\vee\dots\vee s_n\leq p\vee a$ (since every $s_i$ that is not in $I$ must be equal to $a$). Thus $b\leq p\vee a$, as desired.