It is not! In this expression:

$var = func($var) || $var;
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the return value of func($var) will be assigned to $var if func($var) is true, otherwise the value of $var. In this expression

$var ||= func( $var );
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the return value of func($var) will be assigned to $var if $var is false.

Completely different logic!

The expression $var ||= func( $var ); is shorthand for $var = $var || func( $var );. Can you spot the difference?

$var = $var || func( $var );
$var = func( $var ) || $var;
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They're certainly logically equivalent if we disregard the short-circuiting nature of operators. (Though the mathematics branch of logic generally deals with no other return values than true and false...)

As you say, only if we only worry about true or false...

my $var = 5;
sub func { 10 };

print $var || func($var), "\n";
print func($var) || $var, "\n";
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