A finite, cancellative semigroup is a group

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 $a$.

Your three axioms do not describe a group but rather a semigroup with cancellation law. An example of such a beast is the set $\mathbb N$ with addition. It is not a group.

However, you additionally require finiteness. To show that $(G,*)$ is now a group, consider the following: Select $a\in G$ and define $a^n$ recursively: $a^1=a$, $a^{n+1}=a^n*a$. Finiteness implies that $a^n=a^m$ for some $n\ne m$ . Wlog. $n>m$, say $n=m+k$ with $k>0$. Then $a^{k+1}*a^{m}=a^{n+1}=a^n*a=a^m*a=a^{m+1}=a*a^{m}$, hence $a^{k+1}=a$. Let $e=a^{k}$. Then $e*a=a*e=a^{k+1}=a$. Let $x\in G$. Since the $x* y$, $y\in G$ are pairwise distinct, there is one $y$ with $x* y = a$. It follows that $$ (e*x)*y = e*(x*y) = e*a = a = x*y,$$ hence $e*x=x$. Since the $z*x$, $z\in G$ are pairwise distinct, there is one $z$ with $z*x = a$. It follows that $$ z*(x*e)=(z*x)*e=a*e=a=z*x,$$ hence $x*e=x$. Thus $e$ is left and right neutral.

Then, again by the $x* y$ being distinct, we find $y\in G$ such that $x*y=e$. But then also $(y*x)*y=y*(x*y)=y*e=y=e*y$ implies $y*x=e$, i.e. the $y$ found is left and right inverse of $x$.