Showing associativity of (x*y) = (xy)/(x+y+1)

Too long for a comment:

@heropup: Your formula for $x*y$ shows that the $*$ operation is the one obtained by transport of structure of the usual multiplication. Indeed, let $$ f\colon \mathbf R\setminus\{0\}\rightarrow \mathbf R\setminus\{1\} $$ be the map defined by $$ f(x)=1+\frac1x. $$ Then $f$ is a bijection and its inverse map is given by $$ f^{-1}(y)=\frac1{y-1} $$ for all $y\in\mathbf R\setminus\{1\}$. We then indeed have, as per your formula, $$ x*y= \frac{1}{(1+\frac1x)(1+\frac1y)-1}=f^{-1}(f(x)\cdot f(y)), $$ where $\cdot$ is usual multiplication on $\mathbf R\setminus\{1\}$. This is what one calls an operation obtained by transport of structure: if $f\colon E\rightarrow F$ is a bijection between two sets, and $\cdot$ is a binary operation on $F$, then one has an induced binary operation $*$ on $E$ defined by $$ x*y=f^{-1}(f(x)\cdot f(y)) $$ for all $x,y\in E$. Its a kind of ping-pong principle, you map the elements $x$ and $y$ of $E$ in $F$ where you can multiply them, and then map the result back. The operation $*$ is the unique operation on $E$ for which $$ f(x*y)=f(x)\cdot f(y) $$ for all $x,y\in E$. The binary operation $*$ obtained by transport of structure has all properties that the binary operation $\cdot$ has: if $\cdot$ is associative then $*$ is so, if $\cdot$ has a neutral element then $*$ has one, and in that case, if inverses exist for $\cdot$ then inverses exists for $*$, if $\cdot$ has an absorbing element then $*$ has one, if $\cdot$ is commutative then $*$ is so, and so on.

Now, returning to your specific example and the specific bijection $f$, the usual multiplication $\cdot$ on $\mathbf R\setminus\{1\}$ is not an operation on $\mathbf R\setminus\{1\}$. Indeed, the product of two real numbers different from $1$ can very well be equal to $1$, which is not an element of the set $\mathbf R\setminus\{1\}$. Therefore, the formula above only holds for $x,y\in \mathbf R\setminus\{0\}$ for which $f(x)\cdot f(y)\neq1$. That is probably why your operation $*$ has been restricted to the positive real numbers, which corresponds exactly to the subset $(1,+\infty)$ of $\mathbf R\setminus\{1\}$ where $\cdot$ is a binary operation.

In any case, it is now clear that $*$ is associative since $\cdot$ is so, but much more can be derived now: for example, since the product $\cdot$ is a group law on $\mathbf R\setminus\{0\}$, it would be natural to consider $f$ as a map $$ f\colon (\mathbf R\cup\{\infty\})\setminus\{0\}\rightarrow \mathbf R\setminus\{0\} $$ still defined by $f(x)=1+\frac1x$ for $x\in\mathbf R\setminus\{0\}$, and with $f(\infty)=1$. Then $f$ is again a bijection. Since $\cdot$ is a commutative group structure on the codomain of $f$, one deduces that $*$ defines a group structure on $(\mathbf R\cup\{\infty\})\setminus\{0\}$! The neutral element is the element $\infty$. The inverse $x^{*-1}$ of $x$ with respect to $*$ is $$ x^{*-1}=f^{-1}(f(x)^{\cdot-1})= \frac1{\frac1{1+\frac1x}-1}, $$ where $x^{\cdot-1}$ is the usual inverse $\frac1x$.

The moral is that transport of structure is a wonderful machine to create exercises. Pick your favorite group $G$, take any bijection $f\colon E\rightarrow G$, and consider the binary operation on $E$ obtained by transport of structure. Such an operation can be really weird looking, but still you can be sure that it defines a group structure on your set $E$!

The operation is $ x*y = \dfrac{xy}{x+y+1}$, not $\dfrac{x*y}{x+y+1}$.

So $$(x*y)*z = \dfrac{\dfrac{xy}{x+y+1} z}{\dfrac{xy}{x+y+1} + z + 1} = \dfrac{xyz}{xy + xz + yz + x + y + z + 1}$$ and similarly $$ x*(yz) = \dfrac{x\dfrac{yz}{y+z+1}}{x+\dfrac{yz}{y+z+1}+1} = \dfrac{xyz}{xy + xz + yz + x + y + z + 1}$$