Given a group and defined operation. Can't find and prove the that there exist an inverse. Let $G = \{x \in \mathbb{R} \mid x  \ne - 1\}$, and let 
$x * y = x + y + xy$, where $xy$ is a product of $x$ and $y$. We need to show that this is the group.
I had not difficulties with closure axiom, associative law, and identity element, but I can't find the inverse $h \in G$ such that $h*g = e = g*h$, for every $g \in G$.

Since $0$ is also the identity under
$x \ast y = x + y + xy, \tag 1$
$y$ is the inverse of $x$ if
$x + y + xy = 0; \tag 2$
it is also clear that 
$y = 0 \Longleftrightarrow x = 0; \tag 3$
so if 
$x \ne 0, \tag 4$
then 
$y \ne 0, \tag 5$
and thus
$xy \ne 0; \tag 6$
therefore, (2) yields
$\dfrac{1}{y} + \dfrac{1}{x} + 1 = 0, \tag 7$
that is,
$\dfrac{1}{y} = -\dfrac{1}{x} - 1 = -\dfrac{1 + x}{x}, \tag 8$
or
$y = -\dfrac{x}{1 + x}. \tag{10}$
We can easily check this result; indeed, we have
$x -\dfrac{x}{1 + x} - \dfrac{x^2}{1 + x} = \dfrac{x + x^2}{1 + x} - \dfrac{x}{1 + x} - \dfrac{x^2}{1 + x}$
$= \dfrac{x + x^2 - x - x^2}{1 + x} = \dfrac{0}{1 + x^2} = 0. \tag{11}$
