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If there is a standard or well-known way way to accomplish this, someone else will have to provide it. But if you're open to more... _hacky_ alternatives...

Escaping the delimiter character `/` can certainly be a pain, especially when that character is used extensively throughout one or both of the patterns in `sed 's/regex/replacement/'`. That being said, we're not actually limited to using `/`. According to `man sed`, it can be any character (almost, anyway):

```
/regexp/
    Match lines matching the regular expression regexp.  Matching is performed on the current pattern space, which can be modified with commands such as ‘‘s///''.
\cregexpc
    Match lines matching the regular expression regexp.  The c may be any character.
```

So, instead of this:
```bash
sed 's/\/path\/to\/some\/file/\/path\/to\/some\/other\/file/' infile.txt
```
we can instead use:
```bash
sed 's|/path/to/some/file|/path/to/some/other/file|' infile.txt
```

(It's beyond the scope of this answer, but this also works with `sed`'s `y/source/dest/` command, meaning `sed 'y#abc#XYZ#'` would be just as valid as `sed 'y/abc/XYZ/'`)

Additionally, you can force `sed` to do this for only the first line, by prepending a `1` to the command, such as:
```bash
sed '1s/regex/replacement/'
```

I took the liberty of crafting a `sed` command to accomplish the task you described in your question. It matches the first line only if it starts with `#!` and replaces it with `${shebang}`, writing the entirety of the output to STDOUT as normal.
```bash
sed "1s|#!.*|${shebang}|" infile.txt
```

If you want to match the first line unconditionally, then simply remove the `#!.*`. If you want to match a specific subset of shebangs, a more involved regex pattern could be used.