When you install any OS the install program writes a small piece if code to the first sector (the boot sector) of the drive that is set as the boot drive in the bios. From then on, whenever the computer starts up, that small piece of code is loaded into memory and runs. That code tell the computer where the OS (actually the boot loader) is located, that boot loader then runs and the OS starts up.
When you install another, second, OS it's install program sees the previous install and modifies the boot sector code to indicate that there are two OSs. From then on, whenever the computer starts up an "OS choice screen" is displayed that allows you to choose which of the two OSs to boot up. Also, the two OSs are then aware of each other and the drives will retain the same drive letters is both OSs. The only caveat is that the older OS has to be installed before the newer (Win 7 install can't see a Win 10 boot sector and will overwrite whatever is there).
But when you removed the drive befor the second install the install program did not see the other OS and wrote a new boot sector to the (now different) boot drive. So... now you have two different drives, each with a different boot sector code and each code pointing to a different OS, and neither OS is aware that the other exists.
You can mess around with BCDEdit or EasyBCD but that can lead to problems if drive letters are changed (and they will be). An OS won't be able to find any of it's installed programs because the programs will be located on an (as far as that OS is concerned) a different drive, not where it looks for them.
If I were you I'd kill the Win 10 installation with EasyBCD, delete all it's files (ie. it's Windows folder) and use EasyBCD to make the WIn 7 bootable and then (with all the drives connected) install Win 10. There are other workarounds, but they pretty much all result in unintended consequences that are hard to live with.