Well, '*' stands for a binary operation. I am asking this question because I derived the following 2 results:
If $a_{i}$'s form an arithmetic progression then, $$\sum_{i=1}^{n-1} \frac{1}{\sqrt{a_{i+1}}+\sqrt{a_{i}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$$
$$\sum_{i=1}^{n-1} \frac{1}{a_{i+1}a_{i}}=\frac{n-1}{a_{1}a_{n}}$$
I want to ask if the same result(the generalised result) is valid for every binary operation on $f(i)$ and $f(i+1)$ if these form an arithmetic progression?
My answer is that there are two different formulas
the formula you give in the title and the second formula of your text are one thing (proof A below).
The second formula is another (proof B).
The only connection between them being that they are intrinsically due to telescopic cancellation (and the fact that they deal with arithmetic progressions). I don't see any visible composition operation $\star$ generalizing these formulas.
A) Proof of
$$\sum_{i=1}^{n-1} \frac{1}{f(i)f(i+1)}=\frac{n-1}{f(1)f(n)}\tag{1}$$
We start from the LHS of (1), that we transform progressively in order to obtain finally the RHS.
An arithmetic progression can be written under the form $f(i)=ai+b$ for certain (fixed) real numbers $A$ and $B$ ($A$=ratio and $B$=$0$th term) :
With these notations, the LHS of (1) can be written as :
$$LHS=\sum_{i=1}^{n-1} \dfrac{1}{A}\left(\frac{1}{Ai+B}-\frac{1}{A(i+1)+B}\right)$$
By telescopic cancellation :
$$LHS=\dfrac{1}{A}\left(\frac{1}{A+B}-\frac{1}{An+B}\right)$$
Finally :
$$LHS=\dfrac{1}{A}\dfrac{A(n-1)}{(A+B)(An+B)}=\dfrac{(n-1)}{f(1)f(n)}$$
which is the desired RHS.
B)Proof of
$$\sum_{i=1}^{n-1} \frac{1}{\sqrt{a_{i+1}}+\sqrt{a_{i}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}\tag{2}.$$
Here as well, let us start from the LHS. Multiplying numerator and denominator of each fraction by the conjugate quantity of the denominator $\sqrt{a_{i+1}}-\sqrt{a_{i}}$ gives
$$LHS = \sum_{i=1}^{n-1} \frac{\sqrt{a_{i+1}}-\sqrt{a_{i}}}{a_{i+1}-a_{i}}$$
With expression $a_i=Ai+B$,
$$LHS = \sum_{i=1}^{n-1} \frac{\sqrt{a_{i+1}}-\sqrt{a_{i}}}{a}=\frac{1}{a}\sum_{i=1}^{n-1} (\sqrt{a_{i+1}}-\sqrt{a_{i}})$$
finally giving, by telescopic cancellation :
$$LHS =\frac{1}{a}(\sqrt{a_{n}}-\sqrt{a_{1}})$$
It remains to multiply this result by $$\dfrac{\sqrt{a_{n}}+\sqrt{a_{1}}}{\sqrt{a_{n}}+\sqrt{a_{1}}}$$
to obtain the RHS of (2).