Errata

$text =~ s/^\s*$/<p>/mg;

Given $text as "with.\n\n\n\t \nTherefore", this will correctly produce only one "<p>" due to the form of the string.

The general problem can be seen in this Perl program:
my $text = "A\nB\n\nC\n\n\nD\n\n\n\nE\n\n\n\t F";
$text =~ s/^\s*$/<p>/mg;
print $text;

A simple fix* would be to follow the first substitution with:
$text =~ s/<p><p>/<p>/g;
*However, the whitespace before "F" would still be in $text.

The problem of duplicate <p> tags can also be solved with one substitution:

1 while $text =~ s/^\s*$/<p>/m;

^\w+=[^\n\\]*(\\n[^\n\\]*)*

"... disallowed backslashes other than those at the end of lines."

Actually, anything (or nothing) after a "word" followed by an equal sign will match.
The parentheses do not capture anything useful.

These statements also apply to the next regex:

^\w+=([^\n\\]|\\.)* "in a dot-matches-all mode"

It can be modified to require a backslash at the end of continuation lines.
The perl code below disposes of backslashes and backslash-character sequences
after capturing the values in $1.

my $src = 'SRC=array.c\tbuiltin.c math.c \
args.c \ string.c\
time.c';

if ($src =~ m/^\w+=((?:[^\n\\]|\\.)*)$/s)
{
my @v = split /\s+/, $1 =~ s/\\./ /grs;
print "[@v]\n";
}

This one is presented as a faster solution to the continuation-line problem.
The regex captures the values after the equals sign, but again matches any string that starts with \w+=
It is, in effect, a slower and more complex version of (in perl) /^\w+=(.*)/s