Dear Ben, the coupon collector is just running $m$ Poisson processes at the same moment. The coupons of the $j$th type are being collected at rate $p_j\lambda$ and they're totally independent of the other types of the coupons. It's not hard to see why. Imagine you have an infinitesimal period of time ${\rm d}t$. What's the probability that you collect $j$th coupon in it? Well, the probability is $\lambda\,{\rm d} t$ that you collect any coupon, and you still have to multiply it by $p_j$ to be sure that it's the right coupon.
It follows that the probability that he hasn't collected any coupon of the $j$th type by time $t$ is equal to $\exp(-p_j \lambda t)$ - it's exponentially dropping like it always does in the Poisson distribution.
The probability that the guy has collected the coupon of $j$th time by time $t$ is the complement, $1-\exp(-p_j \lambda t)$. Because the coupons of different types are independent throughout the calculation, the probability that he has collected the coupons of all types by time $t$ (or even earlier, at time $T, of course) is simply the product of the last expression over $j$, ${\rm Prob(All\,\,by\,\,}t{\rm)}=\prod_{j=1}^m [1-\exp(-p_j \lambda t)]$ The expectation value of the total number of collected coupons is just $\lambda T$. One doesn't even have to divide the coupons into types here. One returns to the overall Poisson distribution and this is just what the rate means. In a period of time whose length is $T$, the number of events is $\lambda T$ where $\lambda$ is the rate - that's why it's called the rate of the Poisson distribution. Of course, if $T$ is given by a distribution itself, the expectation value of $L$ will be the expectation value of $\lambda T$ because the relationship is linear.