l1 = {{"c", 3.5}, {"a", 2.2}, {"d", 1.2}, {"b", 0.5}};
l2 = {{"a", 1}, {"b", 3}, {"c", 5}, {"d", 9}};
Using the third argument of GroupBy:
f = Values[
Sort@GroupBy[Join[#1, #2], First,
DeleteDuplicates@*Join @@
ReverseSortBy[#, MatchQ[#[[2]], _Integer] &] &]] &;
f[l1, l2]
{{"a", 1, 2.2}, {"b", 3, 0.5}, {"c", 5, 3.5}, {"d", 9, 1.2}}
Or using GroupBy and SortBy:
f = Values@GroupBy[Sort@Join[#1, #2],
First, {#1[[1]], Splice@SortBy[#2, ! IntegerQ[#] &]} & @@ Thread@# &] &;
f[l1, l2]
{{"a", 1, 2.2}, {"b", 3, 0.5}, {"c", 5, 3.5}, {"d", 9, 1.2}}
Or using Association:
sortPair = SortBy[#, ! IntegerQ[#] &] &@*Identity;
assoc = Partition[Sort[Association@*Rule @@@ Join @@ #], Length@#] &;
Merge[assoc@{l1, l2}, sortPair]
(*<|"a" -> {1, 2.2}, "b" -> {3, 0.5}, "c" -> {5, 3.5}, "d" -> {9, 1.2}|>*)
KeyValueMap[{#1, Splice@#2} &]@%
(*{{"a", 1, 2.2}, {"b", 3, 0.5}, {"c", 5, 3.5}, {"d", 9, 1.2}}*)