I don't have a math research job. I don't know everything with certainty. This is just my guesswork that the following is a simplifying approximation of reality.
There's no universal law that a binary operation whose steps include only multiplication and not division of integers cannot be called division. A rational number can be defined as an equivalence class of ordered pairs of integers $(a, b)$ where $b \neq 0$. From now on when I talk about how operations are defined on ordered pairs of integers, I really mean that that's how it's defined on ordered pairs of integers whose second coordinate is nonzero.
For $(a, b)$ is said to be equivalent to $(c, d)$ when ever $ad = bc$. We can do that because it can be shown that that is an equivalence relation. Addition is defined such that the sum of the equivalnce class containing $(a, b)$ and the equivalence class containing $(c, d)$ is the equivalence class containing $(ad + bc, bd)$ and multiplication is defined such that the product of the equivalcne class containing $(a, b)$ and the equivalence class containing $(c, d)$ is the equivalence class containing $(ac, bd)$. It can also be shown that there is a way to define it that way.
Technically, a field is not a set but an ordered triplet of a set and operations on that set. You can denote a set with operations how ever you want and $\mathbb{Q}$ is just one possible variable you could use to denote it. Most mathematicians agree to use the notation $(\mathbb{Q}, +, \times)$ to refer to any field that is isomorphic to the desired structure of the rational numbers with those operations and call it the rational numbers with addition and multiplication. I believe the mathematical community also agreed that once you have a field and you denote the second coordinate for that field as $+$ and the third coordinate of it as $\times$, $a - b$ is defined to mean $a$ plus the additive inverse of $b$ and $a \div b$ is defined to mean $a$ times the multiplicative inverse of $b$ if $b$ is not the additive identity.
The rational numbers the way I defined them with addition and multiplication the way I defined them can be shown to be a field. Let * denote the operation of taking the equivalence class an ordered pair of integers is in. It can be shown that when ever $a$ is an integer and $b$ is a nonzero integer, $*(a, b) = *(a, 1) \div *(b, 1)$. However, this is true only when $a$ is an integer and $b$ is a nonzero integer. Since technically no rational numbers are integers according to this definition, it's in fact the case that $\forall a \in \mathbb{Q}\forall b \in \mathbb{Q}(a, b) \notin \mathbb{Q}$ so it is not correct to say $*(a, b) = *(a, 1) \div *(b, 1)$ or $*(a, b) = a \div b$. Despite that, we can still define division as $*(a, b) \div *(c, d) = *(ad, bc)$ when ever $c$ is nonzero.
Also once $(\mathbb{Q}, +, \times)$ has been defined, $(\mathbb{Z}, +, \times)$ can be redefined in such a way that $\mathbb{Z}$ is a subset of $\mathbb{Q}$; the operations in $(\mathbb{Z}, +, \times)$ are the exact same operations as the ones in $(\mathbb{Q}, +, \times)$ in that order; and $(\mathbb{Z}, +, \times)$ has the exact same structure as it was defined to have when using that to construct $(\mathbb{Q}, +, \times)$. Most authors use the fractional notation synonymously with division so when $a$ an $b$ are elements of set that was originally called $\mathbb{Z}$ with $b$ nonzero, we cannot use $\frac{a}{b}$ to denote $*(a, b)$ but when $a$ and $b$ are elements of $\mathbb{Q}$ with $b$ nonzero, we can use $\frac{a}{b}$ to denote $a \div b$. In some books, the fractional notation is used to denote division when describing the rule for differentiating a quotient.
Source: https://en.wikipedia.org/wiki/Rational_number
