You're right. There is a mathematical explanation for this in addition to the intuitive one you provide: If the null hypothesis is true, the p-value of each test is uniformly distributed between 0 and 1, but the product of p-values will not be uniformly distributed between 0 and 1, so it is not a p-value.
To combine p-values, what we really want is the CDF of the product of p1 and p2 in the distribution formed by the product of uniform random variables. Mathematica can compute this; the general formula is (p1p2)(1-log(p1p2)), and for the two p-values in question this yields p = 0.0019 (which is close to what I said initially, but only because I made a calculation error; .02*.01 = 0.0002). Thanks for pointing out the mistake in my reasoning! I learned something.
To combine p-values, what we really want is the CDF of the product of p1 and p2 in the distribution formed by the product of uniform random variables. Mathematica can compute this; the general formula is (p1p2)(1-log(p1p2)), and for the two p-values in question this yields p = 0.0019 (which is close to what I said initially, but only because I made a calculation error; .02*.01 = 0.0002). Thanks for pointing out the mistake in my reasoning! I learned something.