I now remembered a statement in Grothendieck's Bourbaki talk 190 on faithfully flat descent (Lemma 3 below) which allows to prove:

**Proposition.** *Let $S$ be a locally noetherian scheme. Then any finite, universally scheme-theoretically dominant morphism $S'\to S$ is a strict epimorphism.*

**Remarks.** 1) A finite morphism is an epi iff it is scheme-theoretically dominant. 2) There is a generalization due to Mesablishvili, see anon's answer below.

So the original question has a positive answer provided we strengthen the assumption so as to require it to hold universally i.e. after any base change $T\to S$. Moreover we may also weaken *birational* to *dominant*.

For the proof I shall use three lemmas:

**Lemma 1.** *An epimorphism of schemes which admits a section is strict.*

**Lemma 2.** *Let $u:S'\to S$ be a finite, universally schematically dominant epimorphism of schemes. If $u$ becomes strict after base change by a strict finite epimorphism $v:T\to S$, then $u$ is strict.*

**Lemma 3 (Bourbaki talk 190).** *Let $S$ be a* [EDIT] ~~locally~~ noetherian scheme. Then any finite epimorphism $u:S'\to S$ factors as a composition of a finite number of strict finite epimorphisms.

Let us first give:

**Proof of the Proposition, given the lemmas.** Let $u:S'\to S$ be a finite, universally scheme-theoretically dominant epimorphism with $S$ locally noetherian. By Lemma 3 we can write $u=u_1\dots u_n$ where each $u_i$ is a strict finite epimorphism. By Lemma 2 and induction, it is enough to prove that $u$ is strict after base change by itself. But after such a base change $u$ acquires a section, hence by Lemma 1 it is strict.

Here are now the proofs of the lemmas.

**Proof of Lemma 1.** This is well-known and easy.

**Proof of Lemma 2.** We easily reduce to the case where $S,S'$ are affine, say $S=\text{Spec}(A)$ and $S'=\text{Spec}(A')$. Then we get the diagram:
$$
\begin{array}{ccccc}
A & \stackrel{f}{\to} & A' & \rightrightarrows & A'\otimes_A A' & \\
\downarrow & & \downarrow & & \downarrow & \\
B & \to & B' & \rightrightarrows & B'\otimes_B B' & \simeq (A'\otimes_A A')\otimes_A B\\
\downarrow\downarrow & & \downarrow\downarrow & & & \\
B\otimes_A B & \to & B'\otimes_{A'} B' & & & \\
\end{array}
$$
where $B':=B\otimes_A A'$ and the second row is assumed exact. Since $f$ is universally injective, then the map $B\otimes_A B\to B'\otimes_{A'} B'$ in the last row is injective. Then an easy diagram chase shows that the first row is exact.

**[EDIT]** Below I give the proof of Lemma 3 that was cooked up by myself and user52824 in his comments below. Incorporating the arguments suggested by user52824 required to edit the original proof sketch only in order for the reading to go smoothly. For these reasons, I reckoned that the historical record of the question (and answer) would be just as good if I edited directly in the original text rather than added a patch afterwards. The place where the arguments of user52824 appear is clearly indicated.

**Proof of Lemma 3.** I don't know any place where a proof appears. Here is one. Define $\mathcal{O}_S$-algebras $A_n$ as follows: set $A_0=u_*\mathcal{O}_{S'}$ and
$$
A_{n+1}:=\{a\in A_n\,;\, a\otimes 1=1\otimes a \mbox{ in }
A_n\otimes_{\mathcal{O}_S} A_n\}.
$$
By the very definition, we see that the kernel of the surjective morphism $A_n\otimes_{\mathcal{O}_S} A_n\to A_n\otimes_{A_{n+1}} A_n$ is equal to $0$, since it is generated by the elements of the form $a\otimes 1-1\otimes a$ for $a\in A_{n+1}$. Hence the diagram
$$
A_{n+1} \to A_n \rightrightarrows
A_n\otimes_{\mathcal{O}_S} A_n
\simeq A_n\otimes_{A_{n+1}} A_n
$$
is exact. That is, the morphism of schemes $\text{Spec}(A_n)\to\text{Spec}(A_{n+1})$ is a *strict* finite epimorphism. Assume that the decreasing sequence $A_1\supset A_2\supset\dots$ stabilizes. Then there is $N\ge 0$ such that $A_N=A_{N+1}$. Thus $a\otimes 1=1\otimes a$ in $A_N\otimes_{\mathcal{O}_S} A_N$, for all $a\in A_N$. Hence the multiplication map $A_N\otimes_{\mathcal{O}_S} A_N\to A_N$ is an isomorphism, hence $\mathcal{O}_S\to A_N$ is a finite epimorphism of sheaves of rings, hence an isomorphism and we are done.

It remains to prove that the sequence stabilizes. We follow the hints given by user52824 in the comment below. First note that the formation of $A_n$ commutes with localization, and that the question of stabilization of the sequence of algebras is Zariski-local on $S$. Moreover if for some point $s\in S$ the sequence of germs $A_{n,s}$ stabilizes, then the isomorphism $\mathcal{O}_{S,s}\to A_{n,s}$ extends in a neighbourhood of $s$ and hence we may assume that $S$ is a local scheme with closed point $s$. In particular, we may assume that $S$ (local or not) has finite Krull dimension $d$, and we shall argue by induction on $d$. If $d=0$ then the rings $A_n$ have finite length and hence the sequence must stabilize. If $d\ge 1$ then since $S\setminus \{s\}$ has dimension $<d$ we know that away from $s$, the sequence $A_n$ stabilizes at some $N$. By the same argument as before, we know moreover that away from $s$, the map $\mathcal{O}_S\to A_N$ is an isomorphism. It follows that the $\mathcal{O}_S$-module $A_n/\mathcal{O}_S$ has finite length, for all $n\ge N$. Thus the sequence $A_n/\mathcal{O}_S$ stabilizes, hence also $A_n$ and we are done.

To finish with, I'll make a comment on the nice counterexample given by James. It shows that *dominant* instead of *universally dominant* is not enough. It turns out to be not so easy to prove that the extension $k[x^3,x^5]\to k[x]$ is not universally injective (here is a hint: write $u=x^3$ and $v=x^5$ so $A:=k[x^3,x^5]\simeq k[u,v]/(u^5-v^3)$, and show that after the base change $A\to A/(u-v)$ the nonzero element $c=u^3-u$ maps to $0$). This leads to the question to find nice, easily recognizable classes of (non faithfully flat) universally dominant scheme morphisms. Apart from isolated examples, I must say that I don't have much in stock.