Unless you are doing differential Expression analysis, you should not count the alternate isoform transcripts via htseq-count. The main reason is that htseq-count removes all the ambiguous reads from counting, and any read which are common to two or more transcript will be ambiguous in this case. See how htseq counts the reads in default (union) mode
In your case, gene_A and gene_B become transcript_A and transcript_B of the same gene. That said, of course, you can do it and there is a small note on the same page regarding that:
Can I use htseq-count to count reads mapping to transcripts rather than genes?
In principle, you could instruct htseq-count to count for each of a gene’s transcript individually, by specifying --idattr transcript_id. However, all reads mapping to exons shared by several transcripts will then be considered ambiguous. (See second question.) Counting them for each transcript that contains the exons would be possible but makes little sense for typical use cases. (See first question.) If you want to perform differential expression analysis on the level of individual transcripts, maybe ahve a look at our paper on DEXSeq for a discussion on why we prefer performing such analyses on the level of exons instead.
If you are not doing differential Expression, then you may try featureCounts which is more flexible and extremely fast. http://bioinf.wehi.edu.au/featureCounts/