Let $k$ be an algebraically closed field of char $0$ and suppose $f(y) \in k[y]$ (need not be monic). Let $t$ be an indeterminate and consider the fraction field $k((t))$ of power series ring $k[[t]]$. My question is:
Is $f(y)-t$ irreducible in $k((t))[y]$?
You’re in a situation where your base ring $k[[t]]=\mathfrak o$ is a complete DVR (complete local ring would be enough), and you have a polynomial $F(y)\in\mathfrak o[y]$. When you pass from $\mathfrak o$ to its residue field $\mathfrak o/\mathfrak m=k[[t]]/tk[[t]]=k$, you get a polynomial over $k$, namely $f(y)$. That is, the ring morphism $\mathfrak o\to k$ by $t\mapsto 0$ gives $F\mapsto f$.
So far, so good. I’ve tried to set your problem up so that you see that Hensel’s Lemma may apply. Now, algebraic closedness of $k$ guarantees that all irreducible factors are linear, and in case $f$ factors over $k$ into distinct linear factors, I mean no two are related by a constant factor, then Hensel applies,and you get a factorization of $F=f(y)-t$ into the same number of linear factors. But if there are repeats, maybe not.
Examples: $f(y)=y^n$. Then $y^n-t$ is irreducible, by Eisenstein. On the other hand, consider $y^2+y-t$, say over (the algebraic closure of) $\Bbb Q$. With a handy computation package, I see that the factorization is $(y-r_1)(y+1+r_1)$, where $r_1=t - t^2 + 2t^3 - 5t^4 + 14t^5 - 42t^6 + 132t^7 - 429t^8 +\cdots$