NP-Hard on Complete Graphs
Usually, when reducing problem B to problem A, not all possible instances of problem A appear in the image of the reduction. This is absolutely not an issue.

In your case, you can say that problem A is NP-hard even if restricted only to complete graphs. This is a stronger result, which is a good thing, and definitely allowed.

As an example, 3-coloring is hard even for 4-colorable graphs, and even just for planar graphs. You show this (say) by reduction from SAT to 3-coloring, in which all the resulting graphs are planar. This is not a drawback – in fact it's a bonus.