This is false. I will use freely that $f \colon A \to B$ is an epimorphism if and only if $B \otimes_A B \to B$ is an isomorphism; in particular this is also equivalent to $\operatorname{Spec} B \to \operatorname{Spec} A$ being a monomorphism.

**Example.** Let $k$ be a field of characteristic $0$, and let
$$R = k\left[x,\tfrac{1}{x-1},\tfrac{1}{x-2},\ldots\right].$$
Then $R$ has a self-map $x \mapsto x-1$, whose image on spectra is the complement $U$ of the origin. Note that $U \amalg \{0\} \to \operatorname{Spec} R$ is a monomorphism, so the map
\begin{align*}
g \colon R &\to R \times k\\
x &\mapsto (x-1,0)
\end{align*}
is an epimorphism. Now let $A = R \times k$, and let $f$ be the composition $A \stackrel\pi\twoheadrightarrow R \stackrel g\to A$, where $\pi$ is the natural projection. It is an epimorphism since both $\pi$ and $g$ are, and it is clearly not injective. $\square$

**Picture.** Here is a picture of the situation:

**Remark.** However, the result is true if $A$ has a unique associated prime $\mathfrak p$ (i.e. $A$ is irreducible without embedded primes). Indeed, we get an ascending chain
$$\mathfrak p \subseteq f^{-1}(\mathfrak p) \subseteq f^{-1}(f^{-1}(\mathfrak p)) \subseteq \ldots,$$
which stabilises by the Noetherian hypothesis. But if $\operatorname{Spec} A \to \operatorname{Spec} A$ is a monomorphism, in particular it is injective, so this forces $\mathfrak p = f^{-1}(\mathfrak p)$. Then the base change $A_{\mathfrak p} \to (f_*A)_{\mathfrak p}$ of $f \colon A \to f_*A$ along $A \to A_{\mathfrak p}$ is an epimorphism as well, hence so is the composition $A_{\mathfrak p} \to (f_*A)_{\mathfrak p} \to A_{\mathfrak p}$ since it is a further localisation. But this is an epimorphism of *Artinian* rings, hence surjective by this post. Since both rings are Artinian of the same length, we conclude that it's an isomorphism. Since $\mathfrak p$ is the unique associated prime, the map $A \to A_{\mathfrak p}$ is injective, which gives the result.

I'm not sure what happens if $A$ is allowed to have embedded primes. It seems hard to imagine what a counterexample could look like, but I also have no argument to rule it out.

**Remark.** Note also that there are trivial counterexamples without the Noetherian hypothesis, even if $A$ is a domain. For example, $A = k[x_0,x_1,\ldots]$ has a surjective endomorphism that is not injective.