I believe the issue is with the definition of approximation.
Let's assume that we want to find a subset that sums to $s=0$ (trivial if all numbers are nonnegative). Then the approximate subset sum problem determines whether or not there is a subset within $(1-c)s$, which is just $0$.
Thus "approximate" subset sum is exactly subset sum if there are negative numbers (and thus can't be solved in polynomial time).
You can demand that the target value $C$ is nonzero, but this is insufficient---if $A$ is integral, we can set $C = 1$ and $c = 1/2$ and again our algorithm is forced to obtain an exact answer. Thus, previous work has simply assumed a positive $A$, and we get a nice definition for what we want. Note that if $A$ is positive, the interesting $C$ are very large, whereas with possibly-negative $A$ they may not be.
It is possible (though I'm far from sure) that one can limit $C$ in such a way as to ensure that the problem is interesting even for possibly-negative $A$. Looking at the reduction, it seems that $C >> NM$ may do the trick; particularly if $C$ is exponential in $N$, and $M$ is polynomial (in other words, the negative numbers of $A$ are all very small). But at first glance, this seems to add complexity to potential approximation algorithms without gaining much in return.
As for the reduction: that reduction as-is does not work well if the goal is $C = 0$. Let's say, for example, that you start with a possibly-negative set $A = \{a_1,\ldots,a_n\}$, where the largest negative number is $\geq -M$, for a large positive $M$. We want to find a subset that sums to 0. Then, as per the reduction you cited, we construct $A' = \{a_1 + M,\ldots, a_n + M, M, \ldots, M\}$ with $n$ copies of $M$. We want to find a subset that sums to $NM$. This can trivially be accomplished by taking the $n$ copies of $M$.
Even if $C\neq 0$ this reduction doesn't work quite the way we would (probably) want. We obtain a solution within $(1-c)(NM + C)$. This may be significantly different than a solution within $(1-c)C$.