Suppose $S$ is an $R$-algebra (associative, commutative, with unit...) such that $S$ is free of finite rank $n$ over $R$. Is it necessarily the case that we can find an $R$-basis $y_1, ..., y_n$ for $S$ with $y_1 = 1$?

I can prove this when $n = 2$:

Suppose $x_1, ..., x_n$ is an $R$-basis for $S$, and write $1 = \sum_{i=1}^n a_ix_i$, with $a_i \in R$. We have $S$ faithfully flat over $R$, so we have $(a_1, ..., a_n) = (a_1, ..., a_n)S \cap R = (1)$, so we can find $b_1, ..., b_n$ in $R$ such that $1 = \sum_{i=1}^n a_ib_i$. Now I use the assumption $n = 2$: the matrix $\left( \begin{array}{cc} a_1 & a_2 \\\ -b_2 & b_1 \end{array} \right)$ is unimodular, so $y_1 = a_1x_1+a_2x_2,$ $y_2 = -b_2x_1+b_1x_2$ is an $R$-basis of $S$ with $y_1 = 1.$

More generally, as long as the unimodular row $(a_1, ..., a_n)$ can be completed to a unimodular matrix over $R$, we can use the unimodular matrix to produce an $R$-basis of $S$ with $y_1 = 1$.

Since not every unimodular row can be completed to a unimodular matrix when $n \ge 3$ and $R$ is arbitrary, it seems like it may be possible to construct a counterexample, but I haven't been able to do it (specifying an algebra structure on a rank $3$ $R$-module is surprisingly tricky).