Convergence of power series around another point
Contrary to what hardmath's comment says, your way of expanding out $f(z+w)$ will work.
Assume that $|z| < R-r$. In particular $|z+w| < R$, so the series for $f(z+w)$ converges: \begin{align*} f(z+w) &= \sum_{n=0}^\infty a_n (z+w)^n \\ &= \sum_{n=0}^\infty \sum_{i=0}^n a_n \binom{n}{i} z^i w^{n-i} \end{align*} With proper justification$^{1}$, as the series converges absolutely, we may switch the order of summation: \begin{align*} &= \sum_{i=0}^\infty \sum_{n=i}^\infty a_n \binom{n}{i} w^{n-i} z^i \end{align*} So let $b_i = \sum_{n=i}^\infty a_n \binom{n}{i} w^{n-i}$, and we have what we wanted: $$ f(z+w) = \sum_{i=0}^\infty b_i z^i. $$
$^{1}$Specifically, to justify this, it suffices (see this note) to show that $$\sum_{n=0}^\infty \sum_{i=0}^n |a_n| \binom{n}{i} |z|^i |w|^{n-i} < \infty.$$
But the expression is equal to $$\sum_{n=0}^\infty |a_n| \left(|z| + |w|\right)^n,$$
and since $\big||z| + |w|\big| = |z| + |w| < (R-r) + r = R$, the original series converges absolutely for $|z| + |w|$ which is what the above says.
If we have a function that has been expanded in a power series about the point $x = x_{0}$
\begin{equation} f(x) = \sum\limits_{n=0}^{\infty} a_{n} (x - x_{0})^{n} \end{equation}
and if the radius of convergence of this series is non-zero, then we can generate a power series expansion about a new point $x = x_{1}$ if this point is in the radius of convergence of the original power series. The new series is [1]
\begin{equation} f(x) = \sum\limits_{k=0}^{\infty} b_{k} (x - x_{1})^{k} \end{equation}
\begin{equation} b_{k} = \sum\limits_{n=0}^{\infty} \begin{pmatrix} n+k \\ k \end{pmatrix} a_{n+k} (x_{1} - x_{0})^{n} \end{equation}
[1] Theory and Application of Infinite Series by Konrad Knopp.