您可以使用以下正则表达式：

(?:^|\n)\*\s*Reference:\s*([\s\S]*?)(?=\s*\n\*|$)

匹配：

• (?:^|\n)\*\s*Reference:\s*：* Reference:位于行首
• ([\s\S]*?)：最小字符数，在组1中捕获
• (?=\s*\n\*|$)：空格、换行符和*或行尾的正向前看

regex101上的正则表达式演示

text = `* Agency:  Agency Ni

* Reference: 

https://web.archive.org/web/20230226234118/https://lawfilesext.leg.wa.gov/law/wsr/agency/BuildingCodeCouncil.htm

https://web.archive.org/web/20230303022030/https://lawfilesext.leg.wa.gov/law/wsr/2023/02/23-02-055.htm (#1)

* Citation: WAC 51-52 / WSR 23-02-055
`

reference = text.match(/(?:^|\n)\*\s*Reference:\s*([\s\S]*?)(?=\s*\n\*|$)/)?.[1] ?? ''

console.log(reference)

您可以简单地使用lookarounds：

(?<=Reference: )(.|\n)*?(?=\*)

然后调整输出。
playground的代码示例：

const text = `This is a code update

* Official Name:  Noner

* Pub: https://content.upcodes.co/viewer/washington/wa-mechanical-code-2021

* Agency:  Agency Ni

* Reference: 

https://web.archive.org/web/20230226234118/https://lawfilesext.leg.wa.gov/law/wsr/agency/BuildingCodeCouncil.htm

https://web.archive.org/web/20230303022030/https://lawfilesext.leg.wa.gov/law/wsr/2023/02/23-02-055.htm (#1)

* Citation: WAC 51-52 / WSR 23-02-055

* Draft Doc Title: 

 WSR 23-02-055 (#1)

* Draft Source Doc: https://web.archive.org/web/20230303022030/https://lawfilesext.leg.wa.gov/law/wsr/2023/02/23-02-055.htm (#1)

* Draft Drive: https://drive.google.com/file/d/1pYmwQS3t-ZX-Vyg9yBabtIpXZ7By2G6f/view?usp=share_link ( #1)

* Final Doc Title: 

   IECC Com Update(#1)

   IECC Res Update (#2)

   IECC Res Update (#3)

* Final Source Doc: 
  https://web.archive.org/web/20230303022130/https://apps.leg.wa.gov/wac/default.aspx?cite=51-52&full=true&pdf=true (#1)
 https://web.archive.org/web/20230303022030/https://lawfilesext.leg.wa.gov/law/wsr/2023/02/23-02-055.htm (#2)

* Final Drive: https://web.archive.org/web/20230303022130/https://apps.leg.wa.gov/wac/default.aspx?cite=51-52&full=true&pdf=true (#1)

https://web.archive.org/web/2023030302fdfdfg2130/https://apps.legfdg.gov/wac/default.aspx?cite=51-52&fdsfullfdsf=true&pfdsfdf=true  (#2)

* Effective Date: January 4, 2023`

const patt = /(?<=Reference: )(.|\n)*?(?=\*)/gm
console.log(text.match(patt)?.[0].trim())

请尝试使用此：

const match = description.match(/(?<=\* Reference\s*:)[\s\S]*?\*/)[0].slice(0, -1);
console.log(match)

此RegEx使用您在问题中使用的look behind，但会匹配下一个*字符之前的任何字符（包括换行符）。我使用.slice()来删除最后一个字符，因为RegEx也会匹配最后一个*字符。
我将这段代码与其他答案进行了基准测试，发现它的速度最快，大约快了50%（see benchmark here）。

在Javascript中查看：

(?<=^\* Reference\s*:\s*)\S[^]*?(?=^\s*\*)

说明

• (?<=正后视，Assert从当前位置向左为：
• ^\* Reference\s*:\s*在字符串开头的可选空白字符之间匹配* Reference后跟:
• )关闭后视
• \S匹配非空白字符
• [^]*?匹配包括换行符在内的任何字符，尽可能少
• (?=正向前看，Assert右侧为：
• ^\s*\*匹配字符串开头的可选空白字符，后跟*
• )关闭前瞻

参见regex demo。

const regex = /(?<=^\* Reference\s*:\s*)\S[^]*?(?=^\s*\*)/gm;
const s = `This is a code update

* Official Name:  Noner

* Pub: https://content.upcodes.co/viewer/washington/wa-mechanical-code-2021

* Agency:  Agency Ni

* Reference: 

https://web.archive.org/web/20230226234118/https://lawfilesext.leg.wa.gov/law/wsr/agency/BuildingCodeCouncil.htm

https://web.archive.org/web/20230303022030/https://lawfilesext.leg.wa.gov/law/wsr/2023/02/23-02-055.htm (#1)

* Citation: WAC 51-52 / WSR 23-02-055

* Draft Doc Title: 

 WSR 23-02-055 (#1)

* Draft Source Doc: https://web.archive.org/web/20230303022030/https://lawfilesext.leg.wa.gov/law/wsr/2023/02/23-02-055.htm (#1)

* Draft Drive: https://drive.google.com/file/d/1pYmwQS3t-ZX-Vyg9yBabtIpXZ7By2G6f/view?usp=share_link ( #1)

* Final Doc Title: 

   IECC Com Update(#1)

   IECC Res Update (#2)

   IECC Res Update (#3)

* Final Source Doc: 
  https://web.archive.org/web/20230303022130/https://apps.leg.wa.gov/wac/default.aspx?cite=51-52&full=true&pdf=true (#1)
 https://web.archive.org/web/20230303022030/https://lawfilesext.leg.wa.gov/law/wsr/2023/02/23-02-055.htm (#2)

* Final Drive: https://web.archive.org/web/20230303022130/https://apps.leg.wa.gov/wac/default.aspx?cite=51-52&full=true&pdf=true (#1)

https://web.archive.org/web/2023030302fdfdfg2130/https://apps.legfdg.gov/wac/default.aspx?cite=51-52&fdsfullfdsf=true&pfdsfdf=true  (#2)

* Effective Date: January 4, 2023
`;

const m = s.match(regex);
if (m) console.log(m[0]);

我决定遵循@Nick的想法，不对“subject”字符串做任何假设。
我提出了两种较为宽松的方法，只要它们有效：

• 当没有 Reference 项时（返回空字符串），
• 当 Reference 项具有空内容时，
• 以及当 Reference 项内容在串的末尾（因此后面没有其它项）时。

第一种方法适用于所有情况，无论其内容如何：

let ref_pat = /^\* Reference:\s*(.*\S(?:\s+.*\S)*?)??\s*(?:^\*|(?![\s\S]))/m;
let reference = description.match(ref_pat)?.[1] ?? '';

如果您假设内容不包含星号字符，则可以使用第二种更有效的模式：

let ref_pat = /^\* Reference:\s*([^*]*[^*\s])/m;
let reference = description.match(ref_pat)?.[1] ?? '';

这是唯一的突破，但这一个是从更简单。
无论您选择哪一个，结果都已被修整。