We write $xy$ for $x*y$.
Let $a$ be an element of $G$, and consider the function $f:G \to G$, $f(g)=ag$. This is injective by (2), and so (since $G$ is finite) it must also be surjective. Thus there exists an element $e$ of $G$ such that $f(e)=a$, i.e. $a=ae$.
We want to show that $e$ is in fact a two-sided identity for $G$. So let $b$ be an arbitrary element of $G$, and consider $ab=(ae)b=a(eb)$. Thus by (2), $b=eb$ for all $b$, i.e. $e$ is a left identity. To show $e$ is also a right identity, consider $bb=b(eb)=(be)b$, and apply (3) to conclude that $b=be$.
Now we need to show the existence of inverses. As above, for every $a$ in $G$ the function $g \mapsto ag$ is surjective, so there is an element, $a_R$ say, such that $aa_R=e$. Similarly, there is an element $a_L$ such that $a_La=e$. Since $a_L=a_Le=a_L(aa_R)=(a_La)a_R=ea_R=a_R$, we have that $a_L=a_R$ is a two-sided inverse for $a$.
This doesn't have a complete answer, but maybe it'll help others.
A regular semigroup is a semigroup $S$ such that $I_a = \{ x : axa = a \}$ is non-empty for every $a \in S$. If we set $V(a) = \{ x : axa = a \text{ and } xax= x \}$ then $V(a) \subset I_a$ and $V(a)$ is non-empty iff $I_a$ is non-empty.
An idempotent is an element $a \in S$ such that $aa=a$, and the collection of idempotents of $S$ is denoted $E(S)$.
An inverse semigroup is a regular semigroup in which idempotents commute, or equivalently, such that $V(a)$ consists of a single element for all $a \in S$.
If $axa=a$, then $(axa)x = (a)x$ means that $(ax)(ax) = (ax)$ so that $ax \in E(S)$. Similarly, $x(axa) = x(a)$ means that $xa \in E(S)$.
If $S$ is a group, then $I_a = \{ a^{-1} \}$ and $E(S) = \{1\}$, so $4 \implies 3$.
If $I_a$ is non-empty and $E(S) = \{ e \}$ then there is some $x$ such that $ax = xa = e$. Then $ea =(ax)a = a$ and $ae = a(xa) = a$, so $e$ is a two-sided identity and $x=a^{-1}$ is a two-sided inverse, and $S$ is a group. Hence $3 \implies 4$.
Clearly $4 \implies 1,2$ with $x=a^{-1}$ the unique solution each time.
Assume 1. Then for each $x$ there is a unique $x$ such that $axax=ax$. Note that $a(xax) \in E(S)$ so $x=xax$. Hence $(xa)(xa) = (xax)a = xa$ so $xa \in E(S)$, but also $x(axa) \in E(S)$, so $axa =a$. Hence $x \in I_a$. Now suppose $y \in I_a$. Then $aya=a$, so $(ay)^2 = ayay = ay$ and $ay \in E(S)$. Hence $y=x$, and $1 \implies 2$.
Assume 1. If $a \in S$, let $a^{-1}$ denote the unique solution to $aa^{-1} \in E(S)$. As before, if $aa^{-1}aa^{-1} = aa^{-1}$, so $a^{-1} a a^{-1} = a^{-1}$ and $a^{-1} a a^{-1} a = a^{-1} a$, so $(a^{-1})^{-1} = a$. Hence $a a^{-1} a = a$ as well, and $S$ is a regular semigroup. If $axa=a$, then $axax=ax$ so $x=a^{-1}$, so $S$ also satisfies 2 and the similar $xax=x$, so that $S$ is an inverse semigroup, and by a theorem idempotents commute. That means $(ef)^2 = e^2 f^2 = ef \in E(S)$, but $e^{-1}$ is the unique solution to $ex \in E(S)$. Hence $f=e^{-1}=e$, and the idempotents are unique. Hence $1 \implies 3$.
So we have a single implication remaining: $2 \stackrel{?}{\implies} 1$.
Now something very close to 2 does not imply 4. 2’ Suppose for each $a \in S$ there is exactly one $x$ such that both $axa=a$ and $xax=x$.
Semigroups satisfying $2'$ are called inverse semigroups, and need not be groups. For plain old 2 though, I haven't found a non-group inverse semigroup that works.
Best Answer
Show that for every $x\in G$, there is an $n\in \Bbb N$ such that $x^n$ is idempotent. Then you can claim that for every $x\in G$, some power of $x$ equals $e$.