This is really a response to Karl's beautiful example; I'm posting it as an "answer" only because there isn't enough room to leave it as a comment.
The condition on conductor ideals is one that I had come across by thinking about the dual picture.
Namely, let $f:Y\rightarrow X$ be a finite map of 1-dimensional proper and reduced schemes over an algebraically closed field $k$. Then $Y$ and $X$ are Cohen-Macaulay by Serre's criterion, so the machinery of Grothendieck duality applies. In particular, the sheaves
$f_*O_Y$ and $f_*\omega_Y$ are dual via the duality functor $\mathcal{H}om(\cdot,\omega_X)$, as are
$O_X$ and $\omega_X$. Here, $\omega_X$ and $\omega_Y$ are the ralative dualizing sheaves of
$X$ and $Y$, respectively. Thus, the existence of a trace morphism $f_*O_Y\rightarrow O_X$
is equivalent by duality to the existence of a pullback map on dualizing sheaves $\omega_X\rightarrow f_*\omega_Y$.
In the reduced case which we are in, one has Rosenlicht's explicit description of the dualizing sheaf: for any open $V$ in $X$, the $O_X(V)$-module $\omega_X(V)$ is exactly the set of
meromorphic differentials $\eta$ on the normalization $\pi:X'\rightarrow X$ with the property that $$\sum_{x'\in \pi^{-1}(x)} res_{x'}(s\eta)=0$$
for all $x\in V(k)$ and all $s\in O_{X,x}$.
It is not difficult to prove that if $C$ is the conductor ideal of $X'\rightarrow X$
(which is a coherent ideal sheaf on $X'$ supported at preimages of non-smooth points in $X$),
then one has inclusions
$$\pi_*\Omega^1_{X'} \subseteq \omega_X \subseteq \pi_*\Omega^1_{X'}(C).$$
Since $X'$ and $Y'$ are smooth, so one has a pullback map on $\Omega^1$'s, our question
about a pullback map on dualizing sheaves boils down the following concrete question:
When does the pullback map on meromorphic differentials $\Omega^1_{k(X')}\rightarrow \pi_*\Omega^1_{k(Y')}$ carry the subsheaf $\omega_X$ into $\pi_*\omega_Y$?
By looking at the above inclusions, I was led to conjecture the necessity of conductor ideal containment as in my original post. As Karl's example shows, this containment is not sufficient.
Here is Karl's example re-worked on the dual side:
Set $B:=k[x,y]/(xy)$ and $A:=k[u,v]/(uv)$ and let $f:A\rightarrow B$ be the $k$-algebra map
taking $u$ to $x^2$ and $v$ to $y$. Writing $B'$ and $A'$ for the normalizations, we have
$B'$ and $A'$ as in Karl's example, and the conductor ideals are $(x,y)$ and $(u,v)$.
Now the pullback map on meromorphic differentials on $A'$ is just
$$(f(u)du,g(v)dv)\mapsto (2xf(x^2)dx,g(y)dy).$$
The condition of being a section of $\omega_A$ is exactly
$$res_0(f(u)du)+res_0(g(v)dv)=0,$$
and similarly for being a section of $\omega_B$. But now we notice that
$$res_0(2xf(x^2)dx)+res_0(g(y)dy) = 2 res_0(f(u)du) + res_0(g(v)dv) = res_0(f(u)du)$$
if $(f(u)du,g(v)dv)$ is a section of $\omega_A$. Thus, as soon as $f$ is not
holomorphic (i.e. has nonzero residue) the pullback of the section
$(f(u)du,g(v)dv)$, as a meromorphic differential on $B'$, will NOT lie in the subsheaf $\omega_B$.
Clearly what goes wrong is that the ramification indices of the map $f:A'\rightarrow B'$
over the two preimages of the nonsmooth point are NOT equal. With this in mind, I propose the following addendum to my original number 4):
In the notation of 4) above and of Karl's post, assume that $f'(C_A)=C_B^e$
for some positive integer $e$. Then the trace map $B'\rightarrow A'$ carries $B$
into $A$.
Certainly this rules out Karl's example. I think another way of stating the condition is that the map $f':Spec(B')\rightarrow Spec(A')$ should be "equi-ramified" over the nonsmooth locus of $Spec(A)$, i.e. that the ramification indices of $f'$ over all $x'\in Spec(A')$ which map to the same nonsmooth point in $Spec(A)$ are all equal.
Is this the right condition?
To address Hanno's question about checking that composition gives a graded-commutative ring structure on $End^{*}(\mathbb{Z}) = \oplus_i [\mathbb{Z}, \Omega^{-i} \mathbb{Z}]$ suppose first that
$a \stackrel{f}{\to} b \stackrel{g}{\to} c \stackrel{h}{\to} \Omega^{-1} a$
is a distinguished triangle in the stable category. Then we can produce from this two isomorphic triangles: one by rotation namely
$a \stackrel{-\Omega^{-1}f}{\to} b \stackrel{-\Omega^{-1}g}{\to} c \stackrel{-\Omega^{-1}h}{\to} \Omega^{-1} a$
and one by applying $\Omega^{-1}\mathbb{Z} \otimes_\mathbb{Z}$
$\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} a \stackrel{1_{\Omega^{-1}\mathbb{Z}}\otimes f}{\to} \Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} b \stackrel{\Omega^{-1}\mathbb{Z}\otimes g}{\to}\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} c \stackrel{\Omega^{-1}\mathbb{Z}\otimes h}{\to} \Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z} \Omega^{-1}a$
The point of this is that the natural isomorphism commuting the loops functor across introduces a sign change, which is precisely the one you pick up by changing the order of composition since changing the composition order is equivalent to applying symmetry isomorphisms to the tensor product which is equivalent to commuting loops across.
To be completely explicit about this there are two functors naturally isomorphic to $\Omega^{-1}$ namely $\Omega^{-1}\mathbb{Z}\otimes_\mathbb{Z}(-)$ and $\mathbb{Z}\otimes \Omega^{-1}(-)$ since $\otimes_\mathbb{Z}$ is biexact there are natural transformations commuting $\Omega^{-1}$ with $\otimes_\mathbb{Z}$ namely
$\Omega^{-1}(-)\otimes_\mathbb{Z} (-) \stackrel{\sim}{\rightarrow} \Omega^{-1}((-)\otimes_\mathbb{Z}(-)) \stackrel{\sim}{\leftarrow} (-)\otimes_\mathbb{Z} \Omega^{-1}(-)$
Our example triangles above which can be obtained from one another by first moving the loops to the right and then applying the unit transformation show that there must be a sign attached to this map for this to give an isomorphism of these triangles. In particular for $\otimes_\mathbb{Z}$ to be compatible with the triangulated structure the two natural isomorphisms moving $\Omega$ about must have different signs. These natural isomorphisms are the precise cause of the sign change.
Best Answer
$\newcommand{\C}{\mathbb C} $I think this is OK. The first step is the inclusion of $\C[X,Y]$ into its fraction field which is $\C(X,Y)$. For each irreducible polynomial $f$ (normalised so that the top degree monomial for some ordering is $1$) we map $\C(X,Y)$ to $\C(X,Y)/\C[X,Y]_{(f)}$ and then we map $\C(X,Y)\rightarrow\bigoplus_f\C(X,Y))/\C[X,Y]_{(f)}$ which is the next step in an injective resolution, the kernel of this map is clearly $\C[X,Y]$. Finally, the cokernel of this map is injective (as the global dimension of $\C[X,Y]$ is $2$).
Addendum: A systematic way of getting this resolution as well as identifying the last term is to note that the Cousin complex of $\C[X,Y]$ is an injective resolution (Hartshorne: Residues and duality, SLN 20, p. 239) which in degree $p$ is the sum of the injective hulls of the residue fields of points of dimension $p$.