Identities and inverses in a field are unique

The additive identity is unique: if $0$ and $0^{\mathrm{\prime }}$ both satisfy $x+0=x$ and $x+0^{\mathrm{\prime }}=x$ for all $x$, then $0^{\mathrm{\prime }}=0^{\mathrm{\prime }}+0=0+0^{\mathrm{\prime }}=0$, using that $0$ is an identity, commutativity, and that $0^{\mathrm{\prime }}$ 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, $1^{\mathrm{\prime }}=1^{\mathrm{\prime }}\cdot 1=1\cdot 1^{\mathrm{\prime }}=1$, and of multiplicative inverses: if $xa=1$ and $xb=1$ with $x\mathrm{\ne }0$, then $a=a\cdot 1=a(xb)=(ax)b=1\cdot b=b$ (using $ax=xa=1$).

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