Apart from the lookarounds that you have used, you can try one of the following:

Disable the multiline modifier. InJavascript, by default, ^ matches \A and $ matches \z and use:

^|$

Demo

or

you can capture them in groups as shown below:

(^)[\s\S]*($)

Demo

• (^) - match the start of the first line and capture it in group 1
• [\s\S]* - greedily match 0+ occurences of any character(including newlines)
• ($) - match the end of the last line and capture it in group 2

When you have no need of ^ and $ as line-boundaries, then these symbols act like \A and \z by default. Demo:

let results = 
`a good test with
a second line, and ending with a bad z
a third line, and a bad z
a final line, and a good z`
.match(/^a (\w+)|(\w+) z$/g);

console.log(results);

If you do need to use line-boundary detection in your regex, then you would normally use the /m modifier, and then you can use your idea of look-around.

Alternatively, you could leave out the /m modifier, and just use ^ and $ for \A and \z, but then use look-around for detecting the line boundaries:

• start-of-line: (?<![^\n\r])
• end-of-line: (?![^\n\r])

let results = 
`a bad test with
a good line, and ending with a good z
a third line, after a good line, and a bad z
a final line, and a bad z`
.match(/(?<![^\n\r]). good|good .(?![^\n\r])/g);

console.log(results);

For specific cases you can sometimes use other techniques. For instance, by default (so without /s) the pattern .* will capture anything until the end of the line, so you can be sure to be at a line-end after that capture.