Identities and inverses in a field are unique

The additive identity is unique: if $0$ and $0'$ both satisfy $x + 0 = x$ and $x + 0' = x$ for all $x$, then $0' = 0' + 0 = 0 + 0' = 0$, using that $0$ is an identity, commutativity, and that $0'$ is an identity.

Additive inverses are unique: if $a$ and $b$ both satisfy $x + a = 0$ and $x + b = 0$, then $a = a + 0 = a + (x + b) = (a + x) + b = 0 + b = b$, using associativity and commutativity.

The same two arguments in the abelian group $(F \setminus \{0\}, \cdot)$ give uniqueness of the multiplicative identity, $1' = 1' \cdot 1 = 1 \cdot 1' = 1$, and of multiplicative inverses: if $x a = 1$ and $x b = 1$ with $x \ne 0$, then $a = a \cdot 1 = a (x b) = (a x) b = 1 \cdot b = b$ (using $a x = x a = 1$).

Therefore $0$, $1$ and, for each $x$, its additive inverse $-x$ and (for $x \ne 0$) its multiplicative inverse $x^{-1}$ are the unique elements with their defining properties, so all four notations are well-defined.