Approximate Subset Sum with negative numbers
I believe the issue is with the definition of approximation.

Let's assume that we want to find a subset that sums to $C=0$.  Then the approximate subset sum problem determines whether or not there is a subset within $(1-c)C$, 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 $C &gt;&gt; NM$ may do the trick (if $M$ is the absolute value of the smallest negative number in $A$); 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.  Also, more work would have to be done to make sure this matches the approximate subset sum definition correctly.



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 $[(1-c)(NM + C),NM+C]$.  This may be significantly different than a solution within $[(1-c)C,C]$ for the original problem.  However, as noted above, if $C &gt;&gt; NM$ this is somewhat close to a $(1-c)$ approximation ratio.  We would need to be careful about solutions within $[C,NM+C]$, however.