Recovering Range Parameters

Octave, 82 bytes

function y=f(x)
while any(floor(y=linspace(x(1)+rand,x(end)+rand,numel(x)))-x),end

Running time is non-deterministic, but the code ends in finite time with probability 1.

Try it online!

Explanation

The code defines a function of x that outputs y. The function consists of a while loop.

In each iteration, the right amount (numel(x)) of linearly spaced values are generated (linspace), starting at x(1)+rand and ending at x(end)+rand. These two calls to the rand function give random offsets between 0 and 1, which are applied to the initial and final values of x.

The loop is repeated for as long as any of the floored results differs (-) from the corresponding entry in x.

Python 3, 189 bytes

def f(l):
 R=range(len(l));e=1-1e-9
 for j in R:
  for I in range(j*4):
   i=I//4;L=[((l[i]+I//2%2*e)*(x-j)-(l[j]+I%2*e)*(x-i))/(i-j)for x in R]
   if[x//1 for x in L]==l:return L
 return l

Try it online!

Cubic time.

Has some numerical issues.

R, 86 bytes

function(n){while(any(n-(x=seq(n[1]+runif(1),tail(n,1)+runif(1),l=sum(n|1)))%/%1))0;x}

Try it online!

R port of Luis Mendo's answer; it does issue a number of warnings because of any coercing to logical but these can be ignored.