Proof.
We can find a commutative diagram
\xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \ar[r] & S \ar[d] \\ U \ar[r] & V \ar[r] & W }
with cartesian squares such that U, V, W are of finite type over \mathbf{Z}. Namely, first write S as a cofiltered limit of finite type schemes over \mathbf{Z} with affine transition morphisms using Limits, Proposition 32.5.4 and then descend the morphism X \to Y using Limits, Lemma 32.10.1. This reduces us to the case discussed in the next paragraph.
Assume S is Noetherian. In this case every quasi-coherent ideal is of finite type, hence we do not have to check the condition that S_ i is cut out by a finite type ideal. Set S_0 = S_{red} equal to the reduction of S. Let \eta \in S_0 be a generic point of an irreducible component of S_0. By Noetherian induction on the underlying topological space of S_0, we may assume the result holds for any closed subscheme of S_0 not containing \eta . Thus it suffices to show that there exists an open neighbourhood U_0 \subset S_0 such that the base change f_0 of f to U_0 has property (3).
Let R be a Noetherian domain. Let f : X \to Y be a morphism of finite type schemes over R. By the discussion in the previous paragraph it suffices to show that after replacing R by R_ g for some g \in R nonzero and X, Y by their base changes to R_ g, formation of the scheme theoretic image of f/R commutes with arbitrary base change.
Let Y = V_1 \cup \ldots V_ n be an affine open covering. Let U_ i = f^{-1}(V_ i). If the statement is true for each of the morphisms U_ i \to V_ i over R, then it holds for f. Namely, the scheme theoretic image of U_ i \to V_ i is the intersection of V_ i with the scheme theoretic image of f : X \to Y by Morphisms, Lemma 29.6.3. Thus we may assume Y is affine.
Let X = U_1 \cup \ldots U_ n be an affine open covering. Then the scheme theoretic image of X \to Y is the same as the scheme theoretic imge of \coprod U_ i \to Y. Thus we may assume X is affine.
Say X = \mathop{\mathrm{Spec}}(A) and Y = \mathop{\mathrm{Spec}}(B) and f corresponds to the R-algebra map \varphi : A \to B. Then the scheme theoretic image of f is \mathop{\mathrm{Spec}}(A/\mathop{\mathrm{Ker}}(\varphi )) and similarly after base change (by an affine morphism, but it is enough to check for those). Thus formation of the scheme theoretic image commutes with base change if \mathop{\mathrm{Ker}}(\varphi \otimes _ R R') = \mathop{\mathrm{Ker}}(\varphi ) \otimes _ R R' for all ring maps R \to R'.
After replacing R, A, B by R_ g, A_ g, B_ g for a suitable nonzero g in R, we may assume A and B are flat over R. By Lemma 37.54.3 we may also assume B/A is a flat R-module. Then 0 \to \mathop{\mathrm{Ker}}(\varphi ) \to A \to B \to B/A \to 0 is an exact sequence of flat R-modules, which implies the desired base change statement.
\square
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