No, it absolutely does not help. In the algorithm you are citing, the numbers never leave their binary representation; there is no counting of the units that make up the number. The step of multiplying by 2 is done by shifting the binary representation of the number, which is not the way that this supposed "Ethiopian algorithm" works. The long multiplication algorithm (in binary or not) has complexity O(log(NM)), where N and M are the numbers to be multiplied, while this algorithm has complexity O(NM), and also gets there in a roundabout way.

This algorithm is totally pointless if you're not going to work with a positional notation for the number; I am sure it was never used in the way described, with stones. Anyone capable of understanding the problem in the first place - 34 goats, 7 pieces per goat - would count out the price - here is 7 for the first goat, 7 for the next, etc. - and tell the shaman to take a hike. This story amounts to calling Ethiopians idiots for taking an algorithm suited for positionally represented numbers - or at least numbers in some kind of a concise representation - and using it in a totally unnecessary way with numbers represented as groups of stones, and not understanding the pointlessness of all the drudgery.

No, the algorithm as described is very different from binary multiplication. It requires a number of stones proportional to the result, while the number of bits required when doing binary multiplication is polylogarithmic.

It's a big difference: 1000 * 1000 would require 1000000 stones if done with the "Ehiopian algorithm", but it can be done with less than 1000 stones by using normal binary multiplication.