When dealing with a group, is it possible that we have $\forall _{a\in G}\:\:a^{-1}=fixed$ (and no "individual" invertible elements?) For example if we have $\phi \left(a,b\right)=\left(1+a\right)\left(1+b\right)$ the neutral element is equal to $e = 0$, and if we look for an element $a^{-1}$ for which $a\cdot a^{-1}=e$ I get that $\forall _{a\in G}\:\:a^{-1}=-1$
Is G a group then? If the invertible element is "fixed" for all a?
 A: First of all, $0$ is not the identity of the operation $\phi$ you wrote down. $\phi(a, 0) = (1+a)(1+0) = 1+a$, but this is not $a$ (unless $1=0$ in whatever ring you're defining $\phi$ on).
To get to your general question, no this is not possible except in the trivial group. In a group $G$, every element has a precisely one inverse. That is, if $ab = ac = e$ then $b = c$. This allows us to have a well defined operation $G \longrightarrow G$ via $a \mapsto a^{-1}$. This map is always a bijection, so every element has precisely one inverse that is uniquely determined by that element. In your claimed example, the map $a \mapsto a^{-1}$ has only $-1$ in the image, but for this to be the case that would mean that $|G| = 1$. That would be the case $1=0$ I mentioned above, but ignore that if you're unfamiliar with ring theory. If you meant $\phi$ to be an operation on something like $\mathbb R$ then this is just false.
You can prove that $a \mapsto a^{-1}$ is a bijection by proving that it is its own inverse, i.e. that $(a^{-1})^{-1} = a$.
A: Then for any $a,b \in G$ we have $e = ac = bc$ where $c$ is the fixed inverse. As $c$ is also invertible,
$$\begin{align}
ac = bc &\implies a=b\\
 &\implies |G| = 1 \\
&\implies G = \{e\}.
\end{align}$$
