I'm not entirely sure how this is formally shown. The proposition is that, for the ring of formal power series $R[[x]] = \{\sum\limits_{k=0}^{\infty} r_kx^k | r_k \in R\}$, show that $U(R[[x]]) = \{ u + xf(x) | u \in U(R), f(x) \in R[[x]]\}$.
Any help is appreciated.
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Suppose $f=\sum_{k\ge0}a_kx^k\in R[[x]]$ and that $a_0$ is invertible. You want to find $g=\sum_{k\ge0}b_kx^k$ such that $fg=1$.
Since the coefficient of $x^k$ in $fg$ is given by $$ \sum_{0\le l\le k}a_{k-l}b_l $$ the condition amounts to saying that $a_0b_0=1$ and, for $k>0$. $$ \sum_{0\le l\le k}a_{k-l}b_l=0 $$ that is, $$ b_k=-a_0^{-1}\sum_{0\le l<k}a_{k-l}b_l $$ This implies that, when you have found $b_0,b_1,\dots,b_{k-1}$, you can also determine (uniquely) $b_k$. Therefore $f$ is invertible.
The converse, that is, $f$ invertible implies $a_0$ invertible, is obvious.
Let $f,g$ be two power series, with leading coefficients $a_0$ and $b_0$. It is easy to see the leading coefficient of $fg$ is $a_0b_0$. Therefore, in order to have $fg=1$, you need $a_0b_0=1$, implying $a_0$ is a unit.
On the other hand, suppose $f=u+xh$ for some unit $u$. Then you can show $f$ is a unit in $R[[x]]$ by exhibiting its inverse. Namely, $$ \frac1f = \frac{1}{u+xf'}=u^{-1}\cdot\frac1{1+xhu^{-1}}=u^{-1}\sum_{k=0}^\infty(-xhu^{-1})^k $$ This is using the classic expansion $1/(1+z)=1-z+z^2-z^3+\cdots$.
In general, it is not allowed to take an infinite sum of elements of $R[[x]]$. However, in this case, it is OK. Letting $[x^i]g$ denote the coefficient of $[x^i]$ in $g$, then $$ [x^i]\sum_{k=0}^\infty(-xhu^{-1})^k=\sum_{k=0}^i[x^i](-xhu^{-1})^k $$ The reason that the upper limit of $\infty$ can be replaced with $i$ is because whenever $k>i$, all of the powers of $x$ appearing in $(-xf'u^{-1})^k$ are greater than $i$. Therefore, each coefficient of $\sum_{k=0}^\infty(-xhu^{-1})^k$ is a finite sum, so this infinite sum is well defined.