Uncomment a COBOL program!
COBOL (GnuCOBOL), 191 + 17 = 208 bytes



I "learned" COBOL for this answer, so it's probably not fully golfed.

This is a full program, taking input on what I presume to be standard input and writing to what I presume to be standard output. Perhaps one day I'll return to this and (1) determine whether COBOL has functions and, if so, (2) see whether a function solution would be shorter.

Byte count includes program and compiler flags (-free and -frelax-syntax).

program-id.c.select i assign keyboard line sequential.fd i. 1 l pic X(80). 88 e value 0.open input i perform until e read i end set e to true end-read if not e and l(7:1)&lt;&gt;'*'display l(8:73).


Try It Online

Ungolfed program

program-id. c.

select i assign to keyboard organization line sequential.

fd i.
    1 l pic X(80).
    88 e value 0.

open input i
perform until e
    read i
        end set e to true
    end-read
    if not e and l(7:1) &lt;&gt; '*'
        display l(8:73).


Limitations

The output is, technically speaking, not correct. From my cursory research, it seems the only practical way to store a string in COBOL is in a fixed-size buffer. I have chosen a buffer size of 80 characters, since this is the line length limit for fixed-format programs. This presents two limitations:


Lines longer than 80 characters are truncated.
Lines shorter than 80 characters are right-padded with spaces.


I'm guessing this is acceptable since, well, it's COBOL. If not, I'd be willing to look into alternatives.

Acknowledgments


-166 bytes thanks to Edward H
-2 bytes thanks to hornj