Let $G$ be a non-empty set, $"\cdot"$ be a binary operation on $G$ and a single operation $x \mapsto \bar{x}$ such that $$a\cdot(b \cdot c) = d \cdot (e \cdot c) \implies b = (\bar{a}\cdot d) \cdot e, \forall a,b,c,d,e \in G.$$
Prove that $(G, \cdot)$ is a group.
For $b = c = d = e = a \implies a = (\bar{a} \cdot a ) \cdot a, \forall a \in G$.
For $d = a, e = b \implies b = (\bar{a} \cdot a) \cdot b, \forall a,b \in G$.
Swapping $a$ and $b$ in the hypothesis equation and making $d = b, e = a$ we obtain that $b = (\bar{a} \cdot a) \cdot b$ and $a = (\bar{b} \cdot b) \cdot a$.
I think I need to show that $\bar{\bar{a}} = a, \forall a \in G$ and that $a \cdot \bar{a} = \bar{a} \cdot a = e$ (the identity element), but I don't know how. It would also help If I could prove that the binary operation is associative.
For all $a,b,c,d,e \in G$, $$a\cdot(b \cdot c) = d \cdot (e \cdot c) \implies b = (\bar a\cdot d) \cdot e. \tag 1$$
Lemma 1: For all $x \in G$, we have $\bar x \cdot x$ is a left identity.
Proof: If we take $a=c=d=x$ and $b=e=y$ in $(1)$ we obtain, $$x\cdot(y \cdot x) = x \cdot (y \cdot x) \implies y = (\bar x\cdot x) \cdot y,$$ and $x\cdot(y \cdot x) = x \cdot (y \cdot x)$ is true for all $x,y \in G$, so $y = (\bar x\cdot x) \cdot y$ for all $x,y \in G$. $\qquad \blacksquare$
Lemma 2: For all $x,y,z \in G$, if $x \cdot z = y \cdot z$ then $x=y$, i.e. the operation $\cdot$ is right cancellative.
Proof: If we take $a=b=d=x$, $c=z$ and $e=y$ in $(1)$ we obtain,
$$x\cdot(x \cdot z) = x \cdot (y \cdot z) \implies x = (\bar x\cdot x) \cdot y,$$ and if $x \cdot z = y \cdot z$ then $x\cdot(x \cdot z) = x \cdot (y \cdot z)$, so $x = (\bar x\cdot x) \cdot y = y$. $\qquad \blacksquare$
Lemma 3: For all $x,y \in G$, we have $\bar x \cdot x = \bar y \cdot y$.
Proof: From Lemma 1, we know $\bar x \cdot x$ and $\bar y \cdot y$ are both left identities, so $(\bar x \cdot x) \cdot x = (\bar y \cdot y) \cdot x$, so from Lemma 2 it follows that $\bar x \cdot x = \bar y \cdot y$. $\qquad \blacksquare$
Defintion: Let $1$ denote the element of $G$ that is equal to $\bar x \cdot x$ for all $x \in G$.
Lemma 4: $\bar 1 = 1$.
Proof: By the definition of $1$, since $1 \in G$ we have $1 = \bar 1 \cdot 1$, and from Lemma 1 we know that $1$ is a left identity so $1 = 1 \cdot 1$. Hence $\bar 1 \cdot 1 = 1 \cdot 1$ so, from Lemma 2, we know $\bar 1 = 1$. $\qquad \blacksquare$
Lemma 5: $1$ is a right identity.
Proof: If we take $a=e=1$ and $b=c=d=x$ in $(1)$ we obtain, $$1\cdot(x \cdot x) = x \cdot (1 \cdot x) \implies x = (\bar 1\cdot x) \cdot 1,$$ and $1\cdot(x \cdot x) = x \cdot (1 \cdot x)$ is true for all $x \in G$ since $1$ is a left identity, so $$x = (\bar 1\cdot x) \cdot 1 = (1\cdot x) \cdot 1 = x \cdot 1 \mbox{ for all } x \in G. \qquad \blacksquare$$
Lemma 6: For all $a,b,d,e \in G$, $$a\cdot b = d \cdot e \implies b = (\bar a\cdot d) \cdot e. \tag 2$$
Proof: If we take $c=1$ in $(1)$ we obtain, $$a\cdot(b \cdot 1) = d \cdot (e \cdot 1) \implies b = (\bar a\cdot d) \cdot e,$$ and since $1$ is a right identity $$a\cdot b = d \cdot e \implies a\cdot(b \cdot 1) = d \cdot (e \cdot 1)$$ so the result follows. $\qquad \blacksquare$
Lemma 7: For all $x,y \in G$, we have $\bar x \cdot (x \cdot y) = y$.
Proof: If we take $a=x$, $b=y$, $d=x \cdot y$ and $e = 1$ in $(2)$ we obtain, $$x\cdot y = (x \cdot y) \cdot 1 \implies y = (\bar x\cdot (x \cdot y)) \cdot 1,$$ and $x\cdot y = (x \cdot y) \cdot 1$ since $1$ is a right identity, so $$y = (\bar x\cdot (x \cdot y)) \cdot 1 = \bar x\cdot (x \cdot y). \qquad \blacksquare$$
Lemma 8: For all $x \in G$, we have $\bar{\bar x} = x$.
Proof: If we take $a=\bar x$, $b=x$ and $d=e=1$ in $(2)$ we obtain,
$$\bar x \cdot x = 1 \cdot 1 \implies x = (\bar{\bar x}\cdot 1) \cdot 1,$$ and $\bar x \cdot x = 1 = 1 \cdot 1$ for all $x \in G$, so for all $x \in G$, $$x = (\bar{\bar x}\cdot 1) \cdot 1 = \bar{\bar x}. \qquad \blacksquare$$
Lemma 9: The operation $\cdot$ is associative.
If we take $a = \bar x$, $b = x \cdot (y \cdot z)$, $d = y$ and $e = z$ in $(2)$ we obtain $$\bar x\cdot (x \cdot (y \cdot z)) = y \cdot z \implies x \cdot (y \cdot z) = (\bar{\bar x}\cdot y) \cdot z,$$ and $\bar x\cdot (x \cdot (y \cdot z)) = y \cdot z$ is true for all $x,y,z\in G$ by Lemma 7, so for all $x,y,z\in G$ we have $$x \cdot (y \cdot z) = (\bar{\bar x}\cdot y) \cdot z = (x\cdot y) \cdot z,$$ using Lemma 8. $\qquad \blacksquare$