Some quotes from Wikipedia:
In mathematics, the field with one element is a suggestive name for an object that should behave similarly to a finite field with a single element, if such a field could exist. This object is denoted $F_1$, or, in a French–English pun, Fun. The name "field with one element" and the notation $F_1$ are only suggestive, as there is no field with one element in classical abstract algebra.
The possibility of studying the mathematics of $F_1$ was originally suggested in 1956 by Jacques Tits, published in Tits 1957, on the basis of an analogy between symmetries in projective geometry and the combinatorics of simplicial complexes. $F_1$ has been connected to noncommutative geometry and to a possible proof of the Riemann hypothesis. Many theories of $F_1$ have been proposed, but it is not clear which, if any, of them give $F_1$ all the desired properties.
https://en.wikipedia.org/wiki/Field_with_one_element
The axioms of a field are as follows:
Formally, a field is a set $F$ together with two operations called addition and multiplication. An operation is a mapping that associates an element of the set to every pair of its elements. The result of the addition of $a$ and $b$ is called the sum of $a$ and $b$ and denoted $a + b$. Similarly, the result of the multiplication of $a$ and $b$ is called the product of $a$ and $b$, and denoted $ab$ or $a⋅b$. These operations are required to satisfy the following properties, referred to as field axioms. In these axioms, $a$, $b$ and $c$ are arbitrary elements of the field $F$.
Associativity of addition and multiplication: $a + (b + c) = (a + b) + > c$ and $a · (b · c) = (a · b) · c$.
Commutativity of addition and multiplication: $a + b = b + a$ and $a · > b = b · a$.
Additive and multiplicative identity: there exist two different elements $0$ and $1$ in $F$ such that $a + 0 = a$ and $a · 1 = a$.
Additive inverses: for every $a$ in $F$, there exists an element in $F$, denoted $−a$, called the additive inverse of $a$, such that $a + > (−a) = 0$.
Multiplicative inverses: for every $a ≠ 0$ in $F$, there exists an element in $F$, denoted by $a^{−1}$, $1/a$, or $\frac{1}{a}$, called the multiplicative inverse of $a$, such that $a · a^{−1} = 1$.
Distributivity of multiplication over addition: $a · (b + c) = (a · b) > + (a · c)$.
This may be summarized by saying: a field has two operations, called addition and multiplication; it is an abelian group under the addition, with 0 as additive identity; the nonzero elements are an abelian group under the multiplication, with 1 as multiplicative identity; the multiplication is distributive over the addition.
https://en.wikipedia.org/wiki/Field_(mathematics)
Chaitin's constant is defined as follows:
Let $P_F$ be the domain of a prefix-free universal computable function $F$.
The constant $\Omega_F$ is then defined as $\Omega _{F}=\sum _{p\in > P_{F}}2^{-|p|}$,
where
$\left|p\right|$ denotes the length of a string $p$. This is an infinite sum which has one summand for every $p$ in the domain of $F$. The requirement that the domain be prefix-free, together with Kraft's inequality, ensures that this sum converges to a real number between $0$ and $1$. If $F$ is clear from context then $\Omega_F$ may be denoted simply $\Omega$, although different prefix-free universal computable functions lead to different values of $\Omega$.
https://en.wikipedia.org/wiki/Chaitin%27s_constant
As I understand it the digits of a Chaitin constant are undecidable, hence it is undecidable whether the set $\{\Omega_F\}$ together with the definitions $\Omega_F + \Omega_F = \Omega_F$ and $\Omega_F \centerdot \Omega_F = \Omega_F$ for addition and multiplication respectively, forms a field.
Thus we can assume that it does, taking this assumption as an axiom.
Does the assumed field $\{\Omega_F\}$ meet the requirements for a field with one element, does it have the necessary properties for why such an object was originally suggested?
Just because a number's digits are undecidable doesn't mean every property of it is undecidable. More generally, just because an object $x$ has a property $\phi$ that's undecidable in a given axiom system (i.e., $\phi(x)$ is true in some models and false in others) doesn't mean there isn't a property $\psi$ that is decidable (i.e., $\psi(x)$ is the same in all models).
In particular, for all real numbers $x$ (including all Chaitin numbers), the property $$ \psi(x) = \begin{cases} \mathrm{True} & if\ \ x=x \\ \mathrm{False} & if\ \ x \neq x \end{cases} $$
is decidable. Namely, it's always true. The proof of this depends on what formal system you're working in; often it's just taken as an axiom of the underlying predicate calculus.
As such, the set $\{\Omega_F\}$ that you propose, where $\Omega_F$ is both the additive identity $0$ and the multiplicative identity $1$, does not satisfy the following field axiom for any Chaitin number $\Omega_F$:
$$ \mathrm{F4.}\quad 0 \neq 1. $$
Therefore, it is never a field.