I was looking for a division in a ring of formal power series. Specifically, let be $A$ a commutative ring with unit. Take $A[[x]]$ and $f\in A[x]$ a monic polynomial not invertible in $A[[x]]$ is so defined a division in $A[[x]]$ for f?
for each $h$ in $A[[x]]$ there $\exists!$ r in $A[x]$ and $q \in A[[x]]$ such that h=fg+r and $degr < degf$
for example is trivial is $f(t)=t^N$:
This cannot always be done. I'm afraid I'm nowhere near to getting to the root cause of the problem. Instead I will just show an example.
Let $A=\Bbb{Z}$, $f=x-3$ and $h=1+x+x^2+x^3+\cdots$. If your goal were attainable then there should exist a constant $r\in\Bbb{Z}$ and a power series $$ q(x)=\sum_{i=0}^\infty q_ix^i\in\Bbb{Z}[[x]] $$ such that $$ h=r+(x-3)q(x).\qquad(*) $$ This turns into a system of equations $$ \begin{aligned} 1&=r-3q_0,\\ 1&=q_i-3q_{i+1},&\text{for all $i$.} \end{aligned} $$ So $$ r=1+3q_0=4+9q_1=13+27q_2=\cdots. $$ In other words $$ r=1+3+9+27+\cdots+3^k+3^{k+1}q_k\equiv\sum_{j=0}^k3^j\pmod{3^{k+1}} $$ for all $k$. Therefore $$ -2r=(1-3)r=\cdots\equiv1\pmod{3^{k+1}} $$ for all $k$.
But, $n=1$ is the only integer congruent to $1$ modulo $3^k$ for all $k$. As $-2r\in\Bbb{Z}$, we can conclude that $-2r=1$. As $r$ is also an integer, this is absurd.