By definition, an identity element $e$ satisfies $eg=ge=g$ for all $g\in G$. Hence if $e_1$ is an identity, then $e_1{e}_2=e_2{e}_1=e_1$. And if $e_2$ is an identity, then $e_2{e}_1=e_1{e}_2=e_2$. Hence $e_1=e_2$. Note that this argument makes no use of inverses, and so is valid for monoids.

By definition, an inverse $h$ of $g$ satisfies $hg=gh=e$. So $h_{1}g=g{h}_1=e$ and $h_{2}g=g{h}_2=e$. Hence, on the one hand,

$$h_{1}g{h}_2=\left(h_{1}g\right)h_2=\left(e\right)h_2=h_2$$

and, on the other hand,

$$h_{1}g{h}_2=h_1\left(g{h}_2\right)=h_1\left(e\right)=h_1.$$

Hence $h_1=h_2$.

Yes, this is a group. The identity element is $0$, and inverse is given by $x\mapsto -x$.

No, this is not a group. $0\in R$ has the property that $0\times x=0$ for all real numbers $x$, so it can't be invertible no matter what the identity is.

Yes, this is a group. The identity is $1$, and inverse is given by $x\mapsto \frac{1}{x}$. In fact this group is isomorphic to $(R,+)$; can you name the isomorphism?

Yes, this is a group (in fact isomorphic to $(R,+)$; can you name the isomorphism?). The identity element is $1$, and inverse is given by $x\mapsto 2-x$ (can you explain why, conceptually?).

No, this is not a group. It's easy to be tricked into thinking it is, because if you just work through the algebra, it seems that all of the group axioms hold. However, this operation is not an operation! It's not defined if the denominator is $0$, because then we'd be dividing by zero.

This operation is interesting and useful, though, when it is defined. It shows up in special relativity, where it describes how velocities add relativistically (in units where the speed of light is $1$).