The explanation is given in the following paragraphs. Given a correct input, the LALR parser produces exactly the same sequence of reduce and shift actions as would an LR parser for the same grammar. On an incorrect input, the LALR parser produces the same sequence of actions up to the last shift action, although it might then do a few more reduce actions before reporting the error. So although the LALR parser has fewer states, its behaviour is identical for correct inputs, and extremely similar for incorrect inputs.
The shift error will be caught at exactly the same input point as with the LR parser because the shift actions are precisely the same between the two parsers, which is why the LALR construction cannot introduce a shift-reduce conflict which was not produced by the LR algorithm. A shift of a terminal $A$ can be done in any state whose itemset contains one or more items with the • marker just before $A$. The LR→LALR state merge only merges states with the same itemsets --that is, it merges lookaheads, not itemsets-- so the shift actions available are the same in both state machines.
This is also true of the reduce actions, available if the • is at the end of a production in the itemset. But because lookaheads are taken into account, it's possible that an itemset has more than one reduction action in the LALR state machine. The fact that the lookahead set in the LALR state machine includes tokens merged from different states does not affect the set of tokens which can eventually be shifted; it just means that the LALR state machine will not realise that the token cannot be shifted until it reaches a state in which a reduction is no longer available.