Bash – the meaning of $0 in the Bash shell

The leading dash in the process name (in argv[0], to be precise) is placed there by the process that calls the shell — login or some equivalent. It tells the shell program that it should act as a login shell (read .profile, etc.). The shell program can choose to act as a login shell in other circumstances (for example, bash and zsh act as a login shell if passed the -l command line option).

To answer your questions:

1. The role of - is to tell the shell to behave as a login shell. For bash, passing --login on the command line is equivalent.
2. You can see the role of - on bash called with the --login option — it reads ~/.profile and so on. You can't see -bash if bash was invoked as bash --login rather than -bash because bash wasn't invoked as -bash.
3. Typical shells don't let you choose the zeroth argument, it has to be the command name. With zsh, you can do ARGV0=-bash bash and with ksh93, bash or zsh, you can do (exec -a -bash bash). In mksh or zsh, you can do hash -- -bash=/bin/bash; -bash. You cannot do something like ln -s /bin/bash ./-bash; ./-bash as it's the whole argv[0] that has to start with -. You could do PATH=:$PATH; -bash, but then it has that nasty side effect of adding "" in front of $PATH.
4. Yes.

See also Difference between Login Shell and Non-Login Shell?

Programs like SSH, sudo, su, etc., start the shell using the exec family of system calls. And these system calls allow the calling program to specify every argument, including the function. For example, using sudo to start a login shell:

$ strace -fe execve -o sudo.log sudo -i
root:~ $ echo $0
-bash
root:~ $ logout
$ grep bash sudo.log 
19124 execve("/bin/bash", ["-bash"], [/* 31 vars */]) = 0
19140 execve("/bin/ls", ["ls", "/etc/bash_completion.d"], [/* 40 vars */]) = 0

As can be seen in the first execve call, the array of arguments passed to bash is just ["-bash"] - so argument zero is -bash.