If $a_0+a_1x+\dots$ has a positive radius of convergence $r$, then so does $f(x)= a_1+a_2x+\dots$, in fact, its radius of convergence is also $r$.
By continuity, if $s<r$, then there is a constant $K>0$ (that may very well depend on $s$) such that $|a_1+a_2x+\dots|<K$ as long as $|x|\le s$. This is simply the fact that continuous functions on closed bounded sets are bounded.
Now, $a_0+a_1x+\dots = a_0+xf(x)$. We know that $a_0=1$ since $f(0)=1$. Presumably you want to argue that if $|x|$ is sufficiently small, then $f(x)\ne0$. To do this, you need to argue that $xf(x)\ne -1$.
But if $|x|$ is sufficiently small (say, $|x|\le s$ with $s$ as above), then $|f(x)|<K$. Now, let $\delta>0$ be so small that $\delta<s$, and $\delta K<1$. Then, if $|x|<\delta$, we have $|xf(x)|=|x||f(x)|<\delta K<1$. In particular, this gives that $xf(x)\ne -1$, which (again) I assume is what you really want.