Let $n\in\Bbb N$ be a Natural number with prime factorization
$$n=\prod_{p_i\in \Bbb P} p_i^{e_i}\tag 1$$
Let's suppose that the ordering of the primes is fixed so that $(1)$ is a unique representation of $n$. This means we can identify $n$ uniquely with the sequence of all exponents $e_i$ from $(1)$:
$$\psi(n) := (e_1, e_2, ...) \in S\tag 2$$
where $S$ denotes all integer sequences (or vectors) where only finitely many elements (or components) are non-zero:
$$S=\left\{a\in\Bbb Z^{\Bbb N}\mid \textstyle\sum_{i\in\Bbb N}[a_i\neq0] \in\Bbb N_0 \right\}\tag 3$$
where $[\,\cdot\,]$ denotes the Iverson bracket. The factorization $(1)$ generalizes without problems to positive Rational numbers, and the mapping $\psi:\Bbb Q^+\to S$ induces a group isomorphism
$$(\Bbb Q^+,\cdot) ~\simeq~ (S,+) \tag 4$$
where $+$ in $S$ is defined by component-wise addition.
The next step will be to extend $\psi$ to all of $\Bbb Q$ and to establish an isomorphism of fields. To that end, we have to add two features to $S$:
A finite abelean group $G$ that represents the group of units; $G\simeq\Bbb Z/2\Bbb Z$ for the integer units $\pm1$.
A special element 0 (zero) that's neither an element of $S$ nor of $G$ nor of $S\times G$.
The resulting object be $$S_G := (G\times S)\cup \{0\} \tag 5$$
Now we can extend $\psi$ to all of $\Bbb Q$, so that (4) extends to an isomorphism of fields
$$(\Bbb Q,+,\cdot) \simeq (S_G,\oplus,+)\tag 6$$
Notice that $+$ acts as multiplication in $S_G$! The addition $\oplus$ is implied by the commutative diagram
$$\require{AMScd}\begin{CD} \Bbb Q\times\Bbb Q @>{+}>> \Bbb Q \\ @VV{(\psi,\psi)}V @VV{\psi}V \\ S_G\times S_G @>{\oplus}>> S_G \\ \end{CD}\tag 7$$
i.e. for $a, b\in S_G$:
$$a\oplus b := \psi(\psi^{-1}(a) + \psi^{-1}(b))$$
with $\psi(0) :=0$. For example, the negative $\ominus a$ of an element $a=(g,s) \in S_G\backslash\{0\}$ is $\ominus(g,s) = (g^{-1},s)$.
Question: The operation $+$ of $S$ can be extended to ordinary integer sequences in $\Bbb S=\Bbb Z^{\Bbb N}$ that allow infinitely many elements to be non-zero. Suppose we extend $\Bbb S$ like in $(5)$ to get $\Bbb S_G:=(g\times \Bbb S)\cup\{0\}$. Is there a way to extend $\oplus$ such that $(\Bbb S_G,\oplus,+)$ becomes a field?
What's clear is that $\Bbb S_G$ would be (isomorphic to) a field extension of $\Bbb Q$ of infinite degree because $|\Bbb S_G|=\aleph_1$.
Note: The construction above is not limited to $\Bbb Q$. Instead we could start from some algebraic number field $K$ with ideal class number of 1, i.e. elements of ${\cal O}_K$ have unique factorization into prime elements (up to ordering, and prime elements are only unique up to units in ${\cal O}_K$). So there are many different fields $(S_G,\oplus,+)\simeq K$ to start from when trying to construct $\Bbb S_G$ as of the question. This might require different group of units $G$, like $G=\Bbb Z/4\Bbb Z$ in the case of $K=\Bbb Q(\sqrt{-1})$.
