05AB1E, 17 bytes

0"D34çý.L"D34çý.L

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Explanation:

• 0"D34çý"D34çý: standard 05AB1E quine taken from this 14-bytes answer
• .L: Levenshtein distance with the (implicit) input-string

Note: there are two shorter standard 05AB1E quines, which are 13 bytes each, but unfortunately:

• This one would be longer (20 bytes) with the added .L: 14405833537536D₅BJ.L - try it online;
• This one doesn't work with inputs.

Vyxal D, 10 bytes

`:qp꘍`:qp꘍

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Thanks to Aaron Miller for this one.

`:qp `:qp  # Standard quine
    ꘍    ꘍ # Levenshtein distance

Vyxal D, 13 bytes

`q\`:Ė\`+꘍`:Ė

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`q\`:Ė\`+ `:Ė # Standard eval quine (see https://codegolf.stackexchange.com/a/232956/100664)
         ꘍    # Leveshtein distance with input

10 bytes in Vyxal 2.6, whiich hasn't been released yet

Python 3.8 (pre-release), 135 bytes

exec(s:="l=lambda a,b:min(l(A:=a[1:],B:=b[1:])-(a[0]==b[0]),l(A,b),l(a,B))+1if b>''<a else len(a+b);print(l('exec(s:=%r)'%s,input()))")

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took my solution of levenstein distance (https://codegolf.stackexchange.com/a/233868/103772) and adapted it to a quine. It is very (very) slow but it works

Haskell, 210 bytes

main=interact$show.((x++show x)%);x@(a:c)%y@(b:d)|a==b=c%d|1>0=1+minimum[x%d,c%y,c%d];x%y=length$x++y;x="main=interact$show.((x++show x)%);x@(a:c)%y@(b:d)|a==b=c%d|1>0=1+minimum[x%d,c%y,c%d];x%y=length$x++y;x="

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Computes the levenshtein distance modified from here. (I wrote the code there too) This is combined with a standard quine mechanism. Not much special here.

If you want to play around with it I suggest making edits towards the end of the string. The algorithm is more or less \$O(n+m^4)\$ where \$n\$ is the location of the difference between the input and the source, and \$m\$ is the length of the string after the first difference. So if you add a character to the front of this the compute time is exponential with the length of input and you won't see your result on tio.

Python 2, 278 258 bytes

t=input();s,f='t=input();s,f=%r,lambda m,n:m or n if m*n<1else-~min(f(m-1,n),f(m,n-1),f(m-1,n-1)-((s%%s)[m-1]==t[n-1]));print f(len(s%%s),len(t))',lambda m,n:m or n if m*n<1else-~min(f(m-1,n),f(m,n-1),f(m-1,n-1)-((s%s)[m-1]==t[n-1]));print f(len(s%s),len(t))

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This is just the usual quine in Python, mixed with the Levenshtein algorithm from this answer. Note that it gets quite extremely (thanks to Mr. Xcoder :P) slow.

JavaScript, 113 bytes

This is a valid quine.

f=t=>[...t].map((v,j)=>x=x.map((w,k)=>q=k--?Math.min(q,w,x[k]-(v==u[k]))+1:j+1),x=[...[,...u=`f=${f}`].keys()])|q

f=t=>[...t].map((v,j)=>x=x.map((w,k)=>q=k--?Math.min(q,w,x[k]-(v==u[k]))+1:j+1),x=[...[,...u=`f=${f}`].keys()])|q

console.log(f('f=t=>[...t].map((v,j)=>x=x.map((w,k)=>q=k--?Math.min(q,w,x[k]-(v==u[k]))+1:j+1),x=[...[,...u=`f=${f}`].keys()])|q'));
console.log(f('%'));
console.log(f('12345'));

Idea stolen from other answer.

Python 2 + sequtils, 101 bytes

from sequtils import*;_='from sequtils import*;_=%r;print edit(_%%_,input())';print edit(_%_,input())