Let $S_3$ act on $\Re^3$ by the permutation matrices P(σ), where P(σ) permutes the basis vectors: $$P(σ)(e_i) = e_{σ(i)}$$
Show that, with this formula, $$\sigma(x_1,x_2,x_3) = P(\sigma)(x_1e_1 + x_2e_2 + x_3e_3) = (x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)} )$$
this is an action of $S_3$ on $\Re^3$.
It seems to me that this is not true, since the inverse of a product flips the order. Am I missing something?
Flipping the order is not an issue unless you are insisting that the action be on a specific side. Given any left action of a finite group, you can turn it into a right action by declaring that $x\cdot g = g^{-1}\cdot x$, and vice versa.
Given your specific example, composition of functions permuting indices is a bit weird in terms of the side the action is on. In general, if you define $$\sigma \cdot (x_a,x_b,x_c) = (x_{\sigma(a)},x_{\sigma(b)},x_{\sigma(c)})$$ then it actually follows that $$\sigma\cdot \tau \cdot (x_a,x_b,x_c) = (x_{\tau\sigma(a)},x_{\tau\sigma(b)},x_{\tau\sigma(c)})$$ This is a valid action, but it is not on the left side. It is a right action.