Let me give a longer comment about the more general form of the question suggested in comments.
So let $0 \neq f \in A$ be an element such that $f=a+b$ implies $a$ divides $b$ or $b$ divides $a$. Equivalently, $f=a+b$ implies $a$ divides $f$ or $b$ divides $f$ (this is because $a$ dividing $b$ implies that $a$ divides $f=a+b$, and $a$ dividing $f$ implies $a$ divides $b=f-a$).
This is further equivalent to:
The set $N$ of all non-divisors of $f$ forms an ideal.
The point is that the set of non-divisors of $f$ is always stable under multiplication by elements of $A$, but in the presence of the condition, it is also stable under addition (if $x, y$ do not divide $f$, then neither does $x+y$, since $(x+y)z=f$ implies that either $xz$ or $yz$ divides $f$, so either $x$ or $y$ divides $f$).
$(f)$ contains a unique maximal proper sub-ideal.
To see this, one looks at $A/N$. Here $f$ is nonzero, and all the proper cosets are represented by elements that divide $f$ in $A$. That is, $f$ is in $A/N$ a nonzero element contained in every nonzero principal, hence every nonzero ideal. Thus, the quotient $N+(f)/N \simeq (f) / (f) \cap N$ is simple. The fact that $(f)\setminus (f) \cap N$ consists of elements associated to $f$ (in the sense that they give the same principal ideal) shows that $(f) \cap N$ is a unique such sub-ideal.
A consequence when $f$ is a non-zero divisor:
If such an element exists and is a non-zero divisor, $A$ is a local ring (hence, all the principal ideals contain a unique maximal sub-ideal).
This holds since $f\cdot -:A \rightarrow (f)$ is in this case a module isomorphism.
So in the particular case when $A=R[[X]]$ is a domain, even existence of such nonzero $f$ implies that $R[[X]]$, hence $R$, has to be a local ring.
One final remark: The initial condition, i.e. $N$ forming an ideal, seems a lot stronger than these consequences. I wonder whether among domains $A$, the only example where this can happen are actually valuation rings.