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 $g \in G$.
$G$ is the multiplicative group of nonzero reals with each real $\,r\,$ "renamed" to $\, r-1\,$ via the bijection $\,h': r\to r-1.\,$ Indeed the inverse "unnaming" map $\,h\!:\ r\to r+1\,$ is clearly a homomorphism
$$\overbrace{ab+a+b+1 \,=\, (a+1)(b+1)}^{\ \ \ \textstyle h(a*b)\, =\, h(a)h(b)}$$
$$\begin{align}{\rm thus}\qquad\quad &\ \ \ \ 1_*\ =\ r^{-1}\,*\,r\\[.3em] \iff\ & h(1_*) = h(r^{-1})\,h(r)\\[.3em] \iff\ &\ \ \ \ 1\, =\, (r^{-1}\!+\!1)(r\!+\!1)\\[.3em] \iff\ &\ r^{-1} = \dfrac{1}{r+1}-1 \,=\, \bbox[5px,border:1px solid #c00]{\dfrac{\!\!-r}{r+1}}\end{align}$$
Or $\,\ r^{-1} = h'h(r^{-1}) = h'(h(r)^{-1}) = h'((r\!+\!1)^{-1}) = (r\!+\!1)^{-1}-1$
As above, to perform a group operation $a*b$ in $G,$ unname the operands to reals $\,a,b\to \color{0a0}{ha,hb},\,$ then perform the operation in $\Bbb R$ to get $\color{#0a0}{ha\cdot hb},\,$ then apply $h'$ to rename the result into $G,$ i.e.
$$\begin{align} a*b\,\ &= h'(\color{#0a0}{ha\cdot hb})\\[.3em] \smash{\overset{\large h(\ \ )_{\phantom{|}}\!\!}\iff}\ h(a * b) &=\, h(a)\cdot h(b) \end{align}\qquad$$
As we see above, this is equivalent to the bijection $h$ being a homomorphism. Thus given any bijection on the underlying set of a group we can "transport" the algebraic operations along the bijection to induce the same algebraic structure on the renamed image. The same idea works for any algebraic structure - see transport of structure. Follow the link for more examples.
As you can check, $0$ is the identity element for the operation, since $x*0 = 0*x = x$ for any $x$. Then, given $x\in G$, the inverse of $x$ is an element $y\in G$ such that $$x+y+xy = x*y = 0.$$ Of course, $y = \cfrac{-x}{1+x}$ must be the inverse of $x$ (notice that this is well defined since $x\neq -1$).
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
$x -\dfrac{x}{1 + x} - \dfrac{x^2}{1 + x} = \dfrac{x + x^2}{1 + x} - \dfrac{x}{1 + x} - \dfrac{x^2}{1 + x}$ $= \dfrac{x + x^2 - x - x^2}{1 + x} = \dfrac{0}{1 + x^2} = 0. \tag{11}$
