They agree everywhere on their domain of definition by assumption. Multiplying by (x-1) doesn't change that. What this does change however is that the resulting

They agree everywhere on their domain of definition by assumption. Multiplying by (x-1) doesn't change that. What this does change however is that the resulting functions can be continuously extended to x=1, and because they agree on a dense set (in this case everywhere else) the extended functions must agree as well. In some sense agreeing on a dense set and being continuous are the minimum properties for this to hold, there are alternative ways to reach the same conclusion but those must be based on those or even stronger properties, if the functions are either not continuous or don't agree on a dense set around x=1 the proof fails.