Proof
(related to Proposition: Unique Solvability of $a\ast x=b$ in Groups)
 By hypothesis, $(G,\ast)$ is a group.
 We first show that $a^{1}\ast b$ is the solution of the equation $a\ast x=b.$
 Replacing \(x\) by \((a^{1}\ast b)\) in the equation leads to $a\ast (a^{1}\ast b)=b.$
 Since "$\ast$" is associative, we get $(a\ast a^{1})\ast b=b.$
 Since the inverse element $a^{1}$ is unique, it follows $e\ast b=b,$ where $e$ is a neutral element in $G$ (even "the" neutral element in $G,$ since it is also unique in $G.$)
 It follows $b=b$ which shows that $a^{1}\ast b$ is a solution.
 Now, we show, the solution $a^{1}\ast b$ is unique.
 Assume, $c\in G$ is a solution of the equation $a\ast x=b,$ i.e. $a\ast c=b.$
 Then $a^{1}\ast (a\ast c)=a^{1}\ast b.$
 Since "$\ast$" is associative, we get $(a\ast a^{1})\ast c=a^{1}\ast b.$
 Thus, $e\ast c=a^{1}\ast b.$
 Thus, $c=a^{1}\ast b.$
∎
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983