If you don't know the distribution, the usual approach would be via Taylor expansion.
e.g. see here or top of p 6 here
or
http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables
(You have to recognize that the two sample means are themselves random variables to apply it.)
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Edit:
The formula is directly relevant for your case because $Var(100(y-z)/z) = 100^2 Var(\frac{y}{z} -1) = 100^2 Var(y/z)$.
I don't know of a specific book reference off the top of my head, it feels a bit like asking for a reference for how to do long division.
It's an absolutely standard technique for approximating means and variances, based quite directly (and in a fairly obvious way) off Taylor series, which have been around for 300 years now. It's certainly mentioned in books, but I've never learned it from a book, in spite of encountering it many times - it's always 'expand this transformation in a Taylor series' (usually, but not always about the mean) and 'take expectations' or 'take variances' (or whatever, as necessary).
Once you learn how to do Taylor series (standard early-undergrad mathematics) and know a few properties of expectations and variances (standard early mathematical statistics), you're done; it's something undergrad students are given as an exercise.
I'll see if I can dig up a reference; there's sure to be something in a standard old reference like Cox and Hinkley or Kendall and Stuart or Feller or something (none of which I have to hand at the moment).
"Coefficient of variation" is a statistic that seems to get at what you're describing, where you divide the standard deviation by the mean. However, for your task of saying which group has more variability, it seems straightforward: one group has $100$-times as high a standard deviation of the other. Given that one variable is spread over a range of $200$ and the other over a range of $200,000$, this is what I would expect, and I would be comfortable disagreeing with your assessment that the second group is more variable than the first.
Best Answer
Does 'Coefficient of Variation' fit the bill?
%CV = SD/MEAN x 100%
Your two samples have a %CV of 0.79% and 9.85% respectively. Suppose your expectation was that the %CV would always be below 1%, then 9.85% could be used to trigger an automated action to search for errors. For example, if the large %CV is actually due to a single error, it is easy to identify as the square of its deviation is the largest of any sample.