It seems trivial that ring $R$ itself is a $R$-module. But then can we say R is finitely generated by multiplicative identity? That seems so trivial..
Yes. We can write $R = (1)$, but notice that it's not necessary that every ideal in $R$ is finitely generated:
Choose $R = k[x_1, x_2, \ldots]$ and let $I = (x_1, x_2, \ldots)$. $R$ is finitely generated since $R = (1)$, but $I$ is not finitely generated.