Not quite. Following your method:
$$\dfrac{n!}{(n-r)!\;r!}=\dfrac{(n-1)!}{(n-r)!(r-1)!}+\dfrac{(n-1)!}{(n-1-r)!\;r!}$$
Multiplying all terms by $r!(n-r)!$ (rather than $(r-1)!(n-r)!$)
$$n!=r(n-1)!+(n-1)!(n-r) = r(n-1)!+(n-1)!n - r(n-1)! = n!$$
But it's usually better to start with one side and ending up at the other side. So:
$${{n-1}\choose{r-1}}+{{n-1}\choose {r}} =$$
$$\dfrac{(n-1)!}{(n-r)!(r-1)!}+\dfrac{(n-1)!}{(n-1-r)!\;r!} =$$
$$ \dfrac{r(n-1)!}{(n-r)!r!}+\dfrac{(n-r)(n-1)!}{(n-r)!\;r!} =$$
$$\dfrac{r(n-1)!+n(n-1)!-r(n-1)!}{(n-r)!\;r!} =$$
$$ \frac{n!}{(n-r)!r!} =$$
$${{n}\choose {r}} $$