As others have mentioned, your claim isn't necessarily true. It seems to be true though that we may write $S=1\cdot R \oplus P$ for some projective $R$-module $P$. This shows that $1_S$ can be choosen as a basis element for instance if $R=K[x_1,\ldots,x_n]$ or if $R$ is a PID (basically whenever there is a reason for a projective module to be free).

Let $X_1,\ldots,X_n$ be an $R$-basis of $S$, and write
$$
   S \cong R[x_1,\ldots,x_s]/(x_ix_j -\sum_{k}r_{ijk} x_k,\ a_1x_1+\ldots+a_nx_n-1)
$$
such that sending $X_i$ to $x_i$ is an isomorphism. I claim the ideal
$(a_1,\ldots, a_n)_R$ in equal to $(1)_R$. If that is so, then there are $t_1,\ldots,t_n\in R$ such that $\sum t_i a_i = 1$, and therefore 
$$\varphi: \ S\longrightarrow R: X_i \mapsto t_i$$
is an $R$-module homomorphism mapping $1_S$ to $1_R$. Sending $1_R$ to  $1_S$ defines a splitting of this epimorphism, and therefore $S \cong 1_S\cdot R \oplus {\rm ker}(\varphi)$. So $S$ is a direct sum of $R$ and the projective $R$-module ${\rm ker}(\varphi)$.

All that is left to show is that $(a_1,\ldots,a_n)_R = (1)_R$ actually holds. Assume this is not true, and suppose $a_1,\ldots,a_n$ are all contained in some maximal ideal $\mathfrak m$ of $R$.  Then, by looking at the relation $a_1 x_1+\ldots+a_nx_n-1=0$ mod $\mathfrak m$, we see that $S/\mathfrak m S =0$, which is a contradiction, since $S$ is assumed to be free of rank $n$ over $R$.