Javascript, 228 bytes, score 228

Naive solution I guess:

(r,d)=>{w=0;l=r.length;for(var i=0;i!=-1&&i<l;){i=r.indexOf(d,i);w=w||((i<0)?0:t(r.substr(0,i)+r.substr((i++)+d.length,l),d))}return !r?1:w}
f=a=>{n=a.length;for(p=1;p<=n;p++)for(i=0;i<=n-p;i++)if(t(a,b=a.substr(i,p)))return b}

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Brachylog, 23 bytes, score 23

Rather boring, as it doesn't have a non-trivial eraser.

{|~c₃↺↔At.l>0∧Akc↰}ᶠlᵒh

Try it online! or verify the other test cases (split in two, so TIO doesn't time out.)

How it works

{|~c₃↺↔At.l>0∧Akc↰}ᶠlᵒh
{                 }ᶠ     Find all outputs for the implicit input, that …
 |                        either is itself or …
  ~c₃                     split into three groups in every possible way
                           [,,AAABBB],[,AA,AABB],[AAA,B,BB],[AA,AB,BB] etc. …
     ↺↔                    with the middle moved to the back [AA,BB,AB] …
       A                   assign to A
        t.                 the tail AB is the output of this predicate
          l>0              and its length is greater than 0.
             ∧Ak          Also, [AA,BB] …
                c          concatenated AABB …
                 ↰         used in a recursive call
                           has the same output as this predicate: AB.
                     lᵒh Order all possible erasers by length, take first

Python 3, score ^{... 48 29} 20

+1,eval(bytes([57]))

A completely different approach from the last 2 versions (and also result in manageable generated program). Check out the revision history for some (possibly) interesting ideas.

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Given an arbitrary Python program:

print(1)

it can be transformed, without changing the behavior:

exec("print(1)")

eval(rb'exec("print(1)")')

eval(bytes([101, 120, 101, 99, 40, 34, 112, 114, 105, 110, 116, 40, 49, 41]))

The number list is getting long, so I'll use a shorter list for demonstration. Note that only printable ASCII characters can appear in the string representation and the first character is e, so the first number is 101 and all numbers are at least 32.

eval(bytes([101, 32, 32]))

Rewrite the first number to 57+1+1+1+...+1+1+1, the rest into 9+1+1+...+1; then add some ,9 at the end of the list. This is always possible because of the conditions mentioned before, and the program's behavior is unchanged because the byte 9 is t, a whitespace character.

eval(bytes([57+1+1+...+1, 9+1+1+...+1+1, 9+1+1+...+1, 9, 9,..., 9]))

The number of ,9 added must satisfy that the total number of 9s is equal to the number of +1.

Then replace all the 9 with eval(bytes([57]), and prepend +1,.

+1,eval(bytes([57+1+1+...+1,eval(bytes([57])+1+1+...+1+1,eval(bytes([57])+1+1+...+1,eval(bytes([57]),eval(bytes([57]),...,eval(bytes([57])]))

Done! It's easy to see that this string has the eraser string +1,eval(bytes([57]).

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It's trivial to see that there exists a Python snippet that solves the given problem.

g=lambda s, e: not s or any(
	s[i:].startswith(e) and g(s[:i]+s[i+len(e):], e)
	for i in range(len(s)-len(e)+1))
f=lambda s: min((s[i:j] for j in range(len(s)+1) for i in range(j) if g(s, s[i:j])), key=len)
print(f(input()))

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