Yep, you know what your own mistake was. You have your mole balance right, but your oxidations and reductions are going backwards.
The (I) in "diaminesilver(I) complex" tells you that the oxidation state of silver is +1. This complex is reduced to silver metal. Since the final product is Ag(s), whose oxidation state is 0, the electrons should be on the reactants side; i.e., the silver must be gaining electrons, and not losing them as you have shown.
Similarly, the acetaldehyde is being oxidized, so it is losing electrons; i.e., the e- should appear on the products side.
So the two half-reactions are
Reduction: [Ag(NH3)2](+) + e(-) -> Ag(s) + 2 NH3
Oxidation: CH3CHO + H2O -> CH3COOH + 2H(+) + 2e(-)
So if you add the two half-reactions and remember to form the ammonium at the end like you did before, you'll get the complete answer.