Given the binary operator $*$ of a finite (even small) commutative group, literally or as a table, how can I proceed to identify the name a mathematician knowing group classification would call it (short of learning group classification)?
If there's no silver bullet, what's a standard name for this particular one?
Let $p>2$ be a prime, and integer $a\in[2,p)$ such that $a^{(p-1)/2}\equiv-1\pmod p$. Let $S=\{\infty\}\cup\mathbb F_p$. Define the binary operation $*$ on $S$: $$u*v=\begin{cases} v&\text{if }u=\infty\\ u&\text{if }u\ne\infty\text{ and }v=\infty\\ \infty&\text{if }u\ne\infty\text{ and }v\ne\infty\text{ and }u+v\equiv0\pmod p\\ \displaystyle\frac{u\,v+a}{u+v}\bmod p&\text{otherwise} \end{cases}$$
$(S,*)$ is a commutative group of order $p+1$, with unity $\infty$, and a single other root of unity $0$.
Example for $p=7$, $a=3$ $$\begin{array}{c} &&&&&&&&\quad&\text{order}\\ \infty&0&1&2&3&4&5&6&&1\\ 0&\infty&3&5&1&6&2&4&&2\\ 1&3&2&4&5&0&6&\infty&&8\\ 2&5&4&0&6&3&\infty&1&&4\\ 3&1&5&6&2&\infty&4&0&&8\\ 4&6&0&3&\infty&5&1&2&&8\\ 5&2&6&\infty&4&1&0&3&&4\\ 6&4&\infty&1&0&2&3&5&&8\\ \end{array} $$ Update: Added the order of each element.
Update: The group can be constructed as an Elliptic Curve Group on $\mathbb F_p$ for the curve $y^2=x(x-a)^2$, where $(x,y)=(u^2,u(u^2-a))$. That curve, with the field $\mathbb R$.
Update: Made the unity $\infty$.
Claim. For all $p$ and $a$ as stated above, the group $\newcommand{\P}{\mathbb{P}} \newcommand{\F}{\mathbb{F}} (\P^1(\F_p), *)$ is cyclic, hence is isomorphic to $C_{p+1}$, the cyclic group of order $p+1$.
First, note that we can write the group operation more uniformly using projective coordinates: \begin{align*} (b_0 : b_1) * (c_0 : c_1) = (b_0 c_0 + a b_1 c_1 : b_0 c_1 + b_1 c_0) \, . \end{align*} (Here we have identified $u$ with $(u:1)$ for all $u \in \F_p$ and $\infty$ with $(1:0)$.) Now the condition $a^{(p-1)/2}\equiv-1\pmod p$ means that $a$ is not a square mod $p$, so the field extension $\F_p(\sqrt{a}) \supseteq \F_p$ has degree $2$. We claim that the map \begin{align*} \varphi: \P^1(\F_p) &\overset{\sim}{\to} \F_p(\sqrt{a})^\times/\F_p^\times\\ (b_0 : b_1) &\mapsto \overline{b_0 + b_1 \sqrt{a}} \end{align*} is a group isomorphism. First, $\varphi$ is well-defined as we've quotiented out by $\F_p^\times$ in the codomain: \begin{align*} (\lambda b_0 : \lambda b_1) &\mapsto \overline{\lambda b_0 + \lambda b_1 \sqrt{a}} = \overline{\lambda (b_0 + b_1 \sqrt{a})} = \overline{b_0 + b_1 \sqrt{a}} \end{align*} for any $\lambda \in \F_p^\times$. Since \begin{align*} (b_0 + b_1 \sqrt{a})(c_0 + c_1 \sqrt{a}) = b_0 c_0 + a b_1 c_1 + (b_0 c_1 + b_1 c_0)\sqrt{a} \end{align*} for all $b_0 + b_1 \sqrt{a}, c_0 + c_1 \sqrt{a} \in \F_p(\sqrt{a})^\times$, then $\varphi$ is a group homomorphism. Since $\varphi$ is surjective and $$ \#(\F_p(\sqrt{a})^\times/\F_p^\times) = \frac{p^2-1}{p-1} = p+1 = \#\P^1(\F_p) \, , $$ then $\varphi$ is bijective, hence an isomorphism.
Since every finite subgroup of the group of units of a field is cyclic (see here, for example), then $\F_p(\sqrt{a})^\times$ is cyclic. Then the quotient $\F_p(\sqrt{a})^\times/\F_p^\times$ is also cyclic, and thus $\P^1(\F_p)$ is cyclic as well.
(This answer was inspired by $\S4.2$ of Lemmermeyer's Conics - A Poor Man's Elliptic Curves (Waybackup).)