How to remove punctuation from a string with exceptions using regex in bash

Using the command echo "Jiro. Inagaki' & Soul, Media_Breeze." | tr -d '[:punct:]' prints the string "Jiro Inagaki Soul MediaBreeze".

However, I want to find a regular expression that will remove all punctuation except the underscore and ampersand i.e. I want "Jiro Inagaki & Soul Media_Breeze".

Following advice on character class subtraction from the sources listed at the bottom, I've tried replacing [:punct:] with the following:

• [\p{P}\-[&_]]
• [[:punct:]-[&_]]
• (?![\&_])\p{P}
• (?![\&_])[:punct:]
• [[:punct:]-[&_]]
• [[:punct:]&&[&_]]
• [[:punct:]&&[^&_]]

... but I haven't gotten anything to work so far. Any help would be much appreciated!

Sources:

• Regex: Match any punctuation character except . and _
• https://www.rexegg.com/regex-quickstart.html

You can specify the punctuation marks you want removed, e.g.

>echo "Jiro. Inagaki' & Soul, Media_Breeze." | tr -d "[.,/\\-\=\+\{\[\]\}\!\@\#\$\%\^\*\'\\\(\)]"
Jiro Inagaki & Soul Media_Breeze

Or, alternatively,

>echo "Jiro. Inagaki' & Soul, Media_Breeze." | tr -dc '[:alnum:] &_'
Jiro Inagaki & Soul Media_Breeze