I believe the issue is with the definition of approximation.

Let's assume that we want to find a subset that sums to $s=0$.  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, then if $C = 1$ we can set $c = 1/2$ and again our algorithm is forced to obtain an exact answer. 
In fact, $C$ must be more than polynomial to avoid a contradiction.
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.  This is part of the reason that positive $A$ is a sufficient condition for interesting approximation algorithms.

On the other hand, positive $A$ is probably not quite necessary.  It is possible that one can limit $C$ and $A$ 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 the analysis without gaining much in return.  

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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 $(1-c)C$.