I think this argument works in the case that the inner product is definite (which is the case you are considering anyways). In this case, the Pin group is generated by the set of all normalized non-zero vectors. Therefore, by the universal property of Cl, a rep of Pin gives a unique rep of Cl and restricting this rep gives you back the rep of Pin you started with since they agree on unit vectors. Thus every rep of Pin is the restriction of a rep of Cl. If you know where all the $e_i$'s get mapped, you then get a unique rep of Cl which gives a unique rep of Pin.
If the inner product is not definite you will still get a rep of Cl but I can't see how you know that it will agree with the original rep of Pin.
I think Imaybemissingsomethingbut:ifthisargumentworksinthecasethattheinnerproductisdefinite(whichisthecase you havearepofareconsideringanyways).Inthiscase,the Pin thenyougetarepgroupisgeneratedbytheset of allnormalizednon-zerovectors.Therefore,by the thirdgroupuniversalpropertyofCl, whicharepofPin gives a unique rep of Cl byandrestrictingthisrepgivesyouback the universalpropertyrep of Cl(asMichaelmentionedinhiscomment)Pinyoustartedwithsincetheyagreeonunitvectors. Thus every rep of Pin is the restriction of a unique rep of Cl. If you havetworepsofPinthatagreeonknowwhere all the $e_i$,e_i$'sgetmapped,you then theylifttogetauniquerepofClwhichgivesauniquerepofPin.
If the sameinnerproductisnotdefiniteyouwillstillgeta rep of Cl andsomustbebutIcan'tseehowyouknowthatitwillagreewith the sameoriginalrepof Pinreps.
I think I may be missing something but: if you have a rep of Pin then you get a rep of the third group, which gives a unique rep of Cl by the universal property of Cl (as Michael mentioned in his comment). Thus every rep of Pin is the restriction of a unique rep of Cl.
If you have two reps of Pin that agree on all the $e_i$, then they lift to the same rep of Cl and so must be the same Pin reps.