One of the ways this could fail for $\{x_i\} = \mathbb{Q}$ is if the power series ended up having a finite radius of convergence.
For example, take $y_i = (1+x_i^2)^{-1}$ for all $x_i$. To start, if we're searching for a continuous function $f$ so that $f(x_i) = y_i$ then we are forced to take $f(x) = (1+x^2)^{-1}$ since continuous functions on Hausdorff spaces are uniquely determined by their values on dense subsets of their domains. Now this function happens to be analytic on $|x|<1$, so it has a power series development there,
$ f(x) = \sum_{k=0}^{\infty} a_k x^k. $
However, this power series diverges for $|x| > 1$, so that
$ y_i \neq \sum_{k=0}^{\infty} a_k x_i^k $
for all $x_i$ satisfying $|x_i| > 1$. Likewise, $f$ has a Laurent series for $|x| > 1$ which diverges for $|x| < 1$. There is no power series for $f$ which converges on $\mathbb{R} \setminus \{\pm 1\}$.
Basically, given any $\{y_i\}$, we must at least be able to find a continuous function $f$ such that $y_i = f(x_i)$. Steven's comment gives a clear counterexample. In addition, we obviously want the function to be analytic if we're going to even think about representing it with a power series. And, if we do, this power series must converge on all of $\mathbb{R}$, and hence on all of $\mathbb{C}$. Thus the admissible $\{y_i\}$ you seek are precisely the values on $\mathbb{Q}$ of an entire function $f(x) = \sum_{k=0}^{\infty} a_k x^k$.