There are two uses for '...' - as:

1. part of an expression - see Range Operators
2. statement on its own - equivalent to die "Unimplemented"; - see The Ellipsis Statement

And so...

I don't understand why '...; print 1;' exits after ellipsis and not proceeds to print (v5.28). Can someone explain?

The sequence ...; print 1; does not proceed to print, because it dies before the print, just at the ...;

I understand why 'print ...' breaks: because it makes syntax error.

Yes. Because it is not a statement on its own, it is interpreted as the infix range operator, which lacks arguments - so "syntax error".