Addition modulo $1$ defines a group structure on the halfopen unit interval $[0,1)$. Obviously, this construction does not work if one starts with $[0,1]$ instead of $[0,1)$.
Is it possible to extend addition on $[0,1]$ to a group operation? More precisely:
Does there exist a binary operation $\ast: [0,1]^2 \to [0,1]$ such that $a \ast b = a+b$ whenever $a+b \leq 1$ and such that $([0,1],\ast)$ is a group?
Clearly, if this is the case then $0$ should be the neutral element. However, we did not make much progress beyond this. This question came up when trying to invent exercises for a group theory course. I do not know what the answer is.
There is no such operation, continuous or not. Observe that if $a,b\in [0,1]$ are such that $a+b>1$, then $$ a*b = 1*(a+b-1) .\quad\quad\quad\quad (1) $$
To see this, just write $$ b=(1-a)+(a+b-1)=(1-a)*(a+b-1) ; $$ this gives (1) after multiplication by $a$.
Now fix $a$ and consider $a*x$, $0\le x\le 1$. This defines a bijection on $[0,1]$, and since $0\le x\le 1-a$ gets mapped bijectively onto $[a,1]$, it follows that $x\mapsto 1*(a+x-1)$ must map $(1-a,1]$ bijectively onto $[0,a)$.
Equivalently, $y\mapsto 1*y$ must map $(0,a]$ bijectively onto $[0,a)$, but this isn't working because now the $y$ with $1*y=0$ would have to be in $(0,a]$ for all $a>0$.