Use structural induction on the finite list xs, with the assertion quantified over every accumulator e and every list ys. The equations of the left fold are foldl f e [] = e and foldl f e (x:xs) = foldl f (f e x) xs.
For xs = [], both sides reduce to foldl f e ys. For the constructor case, equational reasoning gives
foldl f e ((x:xs) ++ ys)
= foldl f e (x : (xs ++ ys))
= foldl f (f e x) (xs ++ ys)
= foldl f (foldl f (f e x) xs) ys
= foldl f (foldl f e (x:xs)) ys
The third equality uses the induction hypothesis with accumulator f e x. Hence the stated identity holds. No inspection of ys was required: in Haskell this equality also permits infinite or partially defined ys, with both expressions denoting the same possibly undefined result.

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