Having three integers a,b,n can I prove $n|(a-b)$ by simply proving that $(a-b)\equiv 0\pmod n$.
For example, to prove that $11|(15^{11}-13^{11}-2^{11})$ we can proceed as follows
$$ \begin{split} (15^{11}-13^{11}-2^{11}) &\equiv (15-13-2) \pmod {11} \quad \text{Fermat's little theorem}\\ 15-13-2 &= 0 \end{split} $$
$(15^{11}-13^{11}-2^{11})\equiv 0\pmod {11}$ proves $11|(15^{11}-13^{11}-2^{11})$
Is it correct?
It's perfectly correct; I'd write $15-13-2\equiv0\pmod{11}$ to go along with the argument. It would work the same if, instead of $15$, you have $26$ or $4$.
By the way, you might prove even more: for what positive integer $n$ does $$ 11\mid (15^n-13^n-2^n) $$ hold? Well, this is the same as saying $$ 4^n-2\cdot2^n\equiv0\pmod{11} $$ which becomes $$ 4^n\equiv2\cdot2^n\pmod{11} $$ and, dividing by $2^n$ (which is possible, prove it), $$ 2^n\equiv 2\pmod{11} $$ or again $$ 2^{n-1}\equiv 1\pmod{11} $$ Now $2^2\not\equiv1$ and $2^5\not\equiv1$; therefore the multiplicative order of $2$ modulo $11$ has to be $10$ (it has to be a divisor of $10$). Then we conclude that $n-1$ has to be a multiple of $10$.