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Edited due to mistakes pointed out in the comments:

I think the answer to the problem might be yes. Here are some preliminary thoughts.

First, you know $f_{\nu}(x)$ divides $f(x)$, since $f(x)$ is always in the kernel, the kernel is principal, and $\mathbb{Z}[[x]]$ is a UFD.

Second, you know $f_{\nu}(x)$ is irreducible, else you don't map into a domain.

Third, I think you can probably prove that these polynomials (the $f_{\nu}(x)$) are distinct. If so continue:

Fourth, this all says that $\prod_{\nu}f_{\nu}(x)$ divides $f(x)$ in $\mathbb{Z}[[x]]$. So, to get the result you just have to prove that the constant coefficients agree up to a unit (i.e. up to $\pm 1$). This is equivalent to showing that the constant coefficient of $f_{\nu}(x)$ is $p^{f_{P}\nu_{P}(\alpha)}$ where $P$ is the prime (above $p$) associated to $\nu$ and $f_{P}$ has its usual meaning.

Edited due to mistakes pointed out in the comments:

I think the answer to the problem might be yes. Here are some preliminary thoughts.

First, you know $f_{\nu}(x)$ divides $f(x)$, since $f(x)$ is always in the kernel, the kernel is principal, and $\mathbb{Z}[[x]]$ is a UFD.

Second, you know $f_{\nu}(x)$ is irreducible, else you don't map into a domain.

Third, I think you can probably prove that these polynomials (the $f_{\nu}(x)$) are distinct. If so continue:

Fourth, this all says that $\prod_{\nu}f_{\nu}(x)$ divides $f(x)$ in $\mathbb{Z}[[x]]$. So, to get the result you just have to prove that the constant coefficients agree up to a unit (i.e. up to $\pm 1$). This is equivalent to showing that the constant coefficient of $f_{\nu}(x)$ is $p^{f_{P}\nu_{P}(\alpha)}$ where $P$ is the prime (above $p$) associated to $\nu$ and $f_{P}$ is the inertial degree of $P$ in the ring of integers over $K$. (Sorry for the double use of $f$--for the polynomial and for the inertial degree.)

Okay, I think I finally understand the problem. For some reason I thought we were taking the product over all roots also. Let's try one more time:

The field $\mathbb{Q}(\sqrt{-7})$ has class number $1$. The prime $2$ splits into the distinct primes $(1/2)(-1\pm \sqrt{-7})$.

Let $f(x)=2x^{2}+x+1$. The roots of this polynomial are $(1/4)(-1\pm \sqrt{-7})$, which both have no places $\nu$ where the valuation is positive. (Scratch that. This polynomial is a unit.)===== I'm going to go and think harder about this problem.

Edited due to mistakes pointed out in the comments:

I think the answer to the problem might be yes. Here are some preliminary thoughts.

First, you know $f_{\nu}(x)$ divides $f(x)$, since $f(x)$ is always in the kernel, the kernel is principal, and $\mathbb{Z}[[x]]$ is a UFD.

Second, you know $f_{\nu}(x)$ is irreducible, else you don't map into a domain.

Third, I think you can probably prove that these polynomials (the $f_{\nu}(x)$) are distinct. If so continue:

Fourth, this all says that $\prod_{\nu}f_{\nu}(x)$ divides $f(x)$ in $\mathbb{Z}[[x]]$. So, to get the result you just have to prove that the constant coefficients agree up to a unit (i.e. up to $\pm 1$). This is equivalent to showing that the constant coefficient of $f_{\nu}(x)$ is $p^{f_{P}\nu_{P}(\alpha)}$ where $P$ is the prime (above $p$) associated to $\nu$ and $f_{P}$ has its usual meaning.

As commented below, I had made a false claim. So here is a modification. Let $f(x)=4x^2+9x+4$. The roots are $(1/8)(-9\pm \sqrt{17})$. Note that $\mathbb{Q}(\sqrt{17})$ has class number 1. If we factor $2=p p'$ then the roots looks like $p^2/p'^2, p'^2/p^2$, up to units. In particular, the extensions of the maps $\varphi_{\alpha}$ treat $x$ as an element of valuation $2$. So, any element in the kernel of the extended map (much less a purported generator) must have constant term which has valuation $2$. Thus the constant term is divisible by $4$.

This means that the constant term of $\prod_{\nu}f_{\nu}(x)$ is divisible by $16$, and thus cannot (even up to units) equal $f(x)$.

For a monic example, let $f(x)=x^{2}-5x+2$. The roots are $(1/2)(5\pm \sqrt{17})$ (which are the two prime factors of $2$ in the ring of integers of $\mathbb{Q}(\sqrt{17})$). The extensions of the maps $\varphi_{\alpha}$ treat $x$ as an element of valuation $1$, so anything in the kernel must have constant term divisible by $2$. The product of two such polynomials has constant term divisible by $4$, hence cannot equal $f(x)$ even up to units.

Okay, I think I finally understand the problem. For some reason I thought we were taking the product over all roots also. Let's try one more time:

The field $\mathbb{Q}(\sqrt{-7})$ has class number $1$. The prime $2$ splits into the distinct primes $(1/2)(-1\pm \sqrt{-7})$.

Let $f(x)=2x^{2}+x+1$. The roots of this polynomial are $(1/4)(-1\pm \sqrt{-7})$, which both have no places $\nu$ where the valuation is positive. (Scratch that. This polynomial is a unit.)===== I'm going to go and think harder about this problem.