Let $x * y = \frac {xy}{x + y + 1}.$
$x * y = \frac {xy}{x + y + 1} \neq \frac {x + y + 1}{xy},$ so $*$ is not commutative.
$(x * y) * z = \frac {(xy)z}{x + y + z + 2} = \frac {x(yz)}{z + y + x + 2}$. So, $*$ is associative.
$x * e = \frac {xe}{x + e + 1} = x$, so $e = x + e + 1 = e \text { and } x = -1.$ Further, $\frac {e(-1)}{e + (-1) + 1} = -1.$ So, $*$ has an identity element.
$x * x' = \frac {xx'}{x + x' + 1} = e,$ so $x' = x + x' + 1 = x'.$ Further, $\frac {x'(-1)}{x' + (-1) + 1} = -1.$ Since $e = x + e + 1 = -1,$ every element has an inverse.
Is the above in\correct?
First and foremost as @JonasMeyer pointed out, this operation is not defined on $\Bbb{R}^2$ as you have the set of points $\{(x,y) \in \Bbb{R}^2: x+y=-1 \}$ that make $x \star y$ undefined. This is an important point to address. If you are going on from here, then:
$1.)$ To establish the commutative property, you need to establish that $x\star y = y \star x$. You did correctly write $x\star y = \frac{xy}{x+y+1}$. However, you then flipped the fraction to prove that this operation isn't commutative. Not sure why you did that, but it is not at all the right thing to do. If you evaluate $y \star x$ you get $$y \star x=\frac{yx}{y+x+1} = \frac{xy}{x+y+1} = x \star y$$ hence, the operation is commutative.
$2.)$ In showing the associative property, you would need to establish $(x\star y)\star z= x\star (y\star z)$. In the work you have shown in your post above, you are on the right track, but are missing a step. You should evaluate the following: $$(x\star y)\star z = \frac{(x\star y)z}{(x\star y)+z+1} = \frac{\frac{xy}{x+y+1}z}{\frac{xy}{x+y+1}+z+1}$$ Now you can play around with the algebra and see if you can simplify $\frac{\frac{xy}{x+y+1}z}{\frac{xy}{x+y+1}+z+1}$ to check if it is in fact equal to $x \star (y \star z)$.
$3.)$ For the inverse element $e$, your work above hasn't shown that there is an inverse; you've actually done the opposite. When you set $\frac{xe}{x+e+1} = x$ you inevitably get the equality $$xe = x(x+e+1)$$ We have to be careful here and not divide by $x$ too fast, as @Xoff has pointed out. Otherwise we miss the solution $x=0$. Now that we have caught that case, we assume $x \neq 0$ which makes it safe to divide by $x$. This leads to $x = -1$. So, what you've shown is that if there is an $e$ such that $x \star e = x$ then the only elements that will allow $e$ to function as the identity is $x = -1$ or $x=0$. This actually establishes that there is not an identity element, since there is no $x \neq -1,0$ such that $x \star e = x$.
$4.)$ Without an identity, there is no inverse. This is clear because you need the existence of $x^{-1}$ such that $x \star x^{-1} = e$, and we know from above that $e$ does not exist under this operation.