Let $f(X)=X^3+aX^2+bX+c \in \mathbb Z[X]$ be a polynomial such that $f(X)$ and $f'(X)$ has no common root in $\mathbb C$. Let $\alpha_i$ , $i=1,2,3$ are the distinct roots of $f$ in $\mathbb C$.
Let $D=\prod_{1\le i <j\le 3} (\alpha_i -\alpha_j)^2$ be the discriminant of $f(X)$, so that $D \in \mathbb Z$.
Now let $\phi(X)=X^4-2bX^2-8cX+b^2-4ac=(f'(X))^2-4(a+2X)f(X) \in \mathbb Z[X]$ .
Then does there exist $g(X),h(X) \in \mathbb Z[X]$ such that $f(X)g(X)+\phi(X)h(X)=D$, in $\mathbb Z[X]$ ?
I can see that the gcd of $f(X)$ and $\phi(X)$ in $\mathbb Q[X]$ is $1$. Indeed, let $d \in \mathbb Q[X] $ be the gcd. If $d$ is not a unit, it has a prime factor say $p \in \mathbb Q[X]$ (since $\mathbb Q[X]$ is a PID) . Then $p|f(X)$ and $p|(f'(X))^2$, so $p|f(X), f'(X)$ . Now $f'$ has degree $2$, so $p$ is either a degree $1$ polynomial or an associate of $f'$. But $f,f'$ has no common root , so $p$ cannot be degree $1$, then $p$ is an associate of $f'$ , then $f'|f$ , again contradicting $f$ and $f'$ has no common root. Thus the gcd is 1, hence $\exists A(X), B(X) \in \mathbb Q[X]$ such that $f(X)A(X)+\phi(X) B(X)=1$ . But unfortunately , this doesn't give integer coefficient polynomials.
For your question, we can verify that $g(x)=3x^3-ax^2-5bx+2ab-27c$ and $h(x)=-3x^2-2ax+a^2-4b$ are what you want, so the answer is positive.
Moreover, we may generalize the result slightly.
Proposition. For any $f\in\mathbb{Z}[x]$, there exist $u,v\in\mathbb{Z}[x]$ such that $uf+v(f')^2=\mathrm{Res}(f,f')$.
Proof. For $f(x)=\sum_{i=0}^{n}a_ix^{n-i}$, we have $f'(x)=\sum_{i=0}^{n-1}(n-i)a_ix^{n-i-1}$ and the resultant $\mathrm{Res}(f,f')$ is given by the determinant of the $(2n-1)\times(2n-1)$ matrix $$M=\left(\begin{matrix} a_0 & 0 &\cdots & 0 & na_0 & 0 &\cdots & 0 & 0\\ a_1 & a_0 &\cdots & 0 & (n-1)a_1 & na_0 &\cdots & 0 & 0\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots & \vdots\\ a_{n-2}& a_{n-3}&\cdots &a_0 & 2a_{n-2} & 3a_{n-3}&\cdots&na_0&0\\ a_{n-1}& a_{n-2}&\cdots &a_1 & a_{n-1} & 2a_{n-2}&\cdots&(n-1)a_1&na_0\\ a_n& a_{n-1}&\cdots &a_2 & 0 & a_{n-1}&\cdots&(n-2)a_2&(n-1)a_1\\ 0& a_n&\cdots &a_3 & 0 & 0&\cdots&(n-3)a_3&(n-2)a_2\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots & \vdots\\ 0& 0&\cdots &a_{n-1} & 0 & 0&\cdots&a_{n-1}&2a_{n-2}\\ 0& 0&\cdots &a_n & 0 & 0&\cdots&0&a_{n-1}\\ \end{matrix}\right). $$
Take the $(2n-1)\times(2n-1)$ matrix $$L=\left(\begin{matrix} 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ 0 & 0 &\cdots& 1 & 0 & 0\\ (x^{2n-2})' & (x^{2n-3})' &\cdots& (x^2)' & 1 & 0\\ x^{2n-2} & x^{2n-3} &\cdots& x^2 & x & 1\\ \end{matrix}\right),$$ then $$LM=\left(\begin{matrix} \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots & \cdots\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots & \cdots\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots & \cdots\\ (x^{n-2}f)'& (x^{n-3}f)'&\cdots & (xf)' &f' & (x^{n-1}f')' & (x^{n-2}f')'&\cdots&(xf')'&f''\\ x^{n-2}f& x^{n-3}f&\cdots & xf & f & x^{n-1}f' & x^{n-2}f'&\cdots&xf'&f'\\ \end{matrix}\right).$$
Take another $(2n-1)\times(2n-1)$ matrix $$R=\left(\begin{matrix} 1 \\ -x & 1\\ &\ddots&\ddots\\ & & -x & 1\\ & & & 0 & 1\\ & & & & -x & 1\\ & & & & &\ddots&\ddots\\ & & & & & & -x & 1\\ \end{matrix}\right),$$ which is block diagonal with an order $(n-1)$ block and an order $n$ block. Then $$LMR=\left(\begin{matrix} \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots & \cdots\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots & \cdots\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots & \cdots\\ x^{n-3}f& x^{n-4}f &\cdots & f &f' & x^{n-2}f' & x^{n-3}f'&\cdots& f'&f''\\ 0& 0&\cdots & 0 & f & 0 & 0 &\cdots&0&f'\\ \end{matrix}\right).$$
Hence we have $\mathrm{Res}(f,f')=\det(M)=\det(LMR)=c_1f+c_2f'$, where $c_1,c_2\in\mathbb{Z}[x]$ and $$c_2=\left|\begin{matrix} \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots \\ x^{n-3}f& x^{n-4}f &\cdots & f &f' & x^{n-2}f' & x^{n-3}f'&\cdots& f'\\ \end{matrix}\right|=c_3f+c_4f'$$ for $c_3,c_4\in\mathbb{Z}[x]$. Therefore, $u=c_1+c_3f'$ and $v=c_4$ satisfy $uf+v(f')^2=\mathrm{Res}(f,f')$. $\square$
When $f$ is a monic polynomial of degree $n$, we have $\mathrm{Disc}(f)=(-1)^{n(n-1)/2}\mathrm{Res}(f,f')$. Thus the discriminant $D=\mathrm{Disc}(f)$ can be written as a $\mathbb{Z}[x]$-combination of $f$ and $(f')^2$, and it is easy to transfer to a $\mathbb{Z}[x]$-combination of $f$ and $\phi=(f')^2-qf$ for any $q\in\mathbb{Z}[x]$. This answers your question in a general aspect.