Here's a definition (taken from here, p.9):
The $\leq$ relation is a partial order relation that stands for 'part of'. E.g. $x\leq x\oplus y$.
In another source, the same author illustrates this definition as follows (p. 21): if $f=\{(x,a),(x\oplus y,b)\}$ then $\ast f=\{(x,a),(x\oplus y,a\oplus b)\}$. I don't see how he obtained $\ast f$. According to the definition above, $\ast f$ should map $x$ to $\bigoplus\{f(x)\}$, which is $\bigoplus \{a\}$, which is $\{a\}$ (by the definition of $\bigoplus$ for sets). As for $x\oplus y$, its image under $\ast f$ should be $\bigoplus\{f(x),f(x\oplus y)\}=\bigoplus \{a,b\}=\{a,b, a\oplus b\}$. So $\ast f$ should be the set consisting of the two pairs $(x,\{a\})$ and $(x\oplus y,\{a,b, a\oplus b\}$). Where is $\ast f=\{(x,a),(x\oplus y,a\oplus b)\}$ coming from?
A heuristic point to keep in mind is that formal semantics is pegged to the realm of natural language, not to the realm of mathematics. So, the mathematical notation may appear sloppy at times, and clear examples from natural language may helps us better to work through the issues.
Lest a function be undefined in some cases, Champollion resorts to the notion of partial function. The definition tells us that $f$'s algebraic closure, $^\ast f$, is the partial function that maps any sum of things $x$ contained in the domain of $f$ to the $\oplus$-sum of their values.
We have $f=\{\langle x,a\rangle,\langle x\oplus y, b\rangle\}$. Let us see by what possibilities $z$ can be instantiated:
For $x$, $x\leq x\wedge a=f(x)$, hence we have $\langle x, a\rangle$ in $^\ast f$.
For $x\oplus y$, $z$ can be $x$, $y$, $x\oplus y$:
$x\leq x\oplus y\wedge a=f(x)$.
$y\leq x\oplus y\wedge f(y)$ is undefined.
$x\oplus y\leq x\oplus y\wedge b=f(x\oplus y)$.
Their sum is $a\oplus b$, hence we have $\langle x\oplus y,a\oplus b\rangle$ in $^\ast f$.
There remains only the combination $\langle x, a\rangle\oplus\langle x\oplus y,a\oplus b\rangle$ as possible.
Using associativity, $\langle (x\oplus x)\oplus y, (a\oplus a)\oplus b\rangle$, and then, idempotence, we get $\langle x\oplus y, a\oplus b\rangle$ again. Therefore,
$^\ast f=\{\langle x,a\rangle,\langle x\oplus y, a\oplus b\rangle\}$.
See also my post on the same subject, linguistic application of mereology.