There is no such thing as "injective" or "surjective" morphisms; these conditions only make sense in the presence of a forgetful functor to $\text{Set}$.
What is true is that exact functors preserve monomorphisms and epimorphisms, and this is because
- in an abelian category, monomorphisms are precisely the morphisms with zero kernel and epimorphisms are precisely the morphisms with zero cokernel,
- exact functors preserve kernels and cokernels, and
- exact functors preserve zero.
Edit: The answer to the revised question is no. The next few paragraphs explain an abstract argument for finding large classes of counterexamples, and towards the end give a specific counterexample, which is easier to understand.
For starters, let $F$ be the identity functor between two copies of the same abelian category $A$ which are concretized with respect to two different forgetful functors $U_1, U_2 : A \to \text{Set}$. If $f$ is a morphism in $A$, call it $U_1$-surjective if $U_1(f)$ is surjective and $U_2$-surjective if $U_2(f)$ is surjective. We are looking for examples of $A, U_1, U_2$ such that $U_1$-surjectivity does not imply $U_2$-surjectivity, which will turn out to be a bit easier to find than examples involving injectivity.
However, if $F : C \to D$ is a counterexample to the question about surjectivity, then $F^{op} : C^{op} \to D^{op}$ is a counterexample to the question about injectivity, provided that the forgetful functors $U_C, U_D : C, D \to \text{Set}$ are replaced by the composites of their opposites
$$U_C^{op}, U_D^{op} : C^{op}, D^{op} \to \text{Set}^{op}$$
with the functor $\text{Hom}(-, 2) : \text{Set}^{op} \to \text{Set}$.
We can construct a general class of counterexamples as follows. Suppose $U_1, U_2$ take the form $\text{Hom}(u_1, -)$ and $\text{Hom}(u_2, -)$ for two generators $u_1, u_2$ of $A$, and that $u_1$ is projective but $u_2$ is not. This means that $U_1$ is exact and faithful, and hence that a morphism is $U_1$-surjective iff it's an epimorphism (exercise).
It also means that $u_2$ admits a nontrivial extension
$$0 \to a \to b \xrightarrow{f} u_2 \to 0$$
for some objects $a, b \in A$. The morphism $f$ is an epimorphism, hence $U_1$-surjective, but it cannot be $U_2$-surjective: applying $\text{Hom}(u_2, -)$ to this short exact sequence produces the Ext long exact sequence
$$0 \to \text{Hom}(u_2, a) \to \text{Hom}(u_2, b) \xrightarrow{U_2(f)} \text{Hom}(u_2, u_2) \xrightarrow{\partial} \text{Ext}^1(u_2, a) \to \cdots$$
where $U_2(f)$ cannot be surjective because the connecting homomorphism $\partial$ is not zero: it necessarily maps $\text{id} \in \text{Hom}(u_2, u_2)$ to the element of $\text{Ext}^1(u_2, a)$ classifying the above extension. From here it suffices to find an abelian category $A$ and two generators $u_1, u_2$ of it, one of which is projective and one of which is not.
This state of affairs is easy to arrange; arguably the simplest example is $A = \text{Ab}, u_1 = \mathbb{Z}, u_2 = \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, so that
$$U_1(-) = \text{Hom}(\mathbb{Z}, -) : \text{Ab} \to \text{Set}$$
is the usual forgetful functor, but
$$U_2(-) = \text{Hom}(\mathbb{Z}, -) \oplus \text{Hom}(\mathbb{Z}/2\mathbb{Z}, -) : \text{Ab} \to \text{Set}$$
sends an abelian group to its underlying set times its subgroup of elements of order dividing $2$. So a morphism $f : x \to y$ of abelian groups is $U_1$-surjective iff it's surjective, but $U_2$-surjective iff it is both surjective and surjective on elements of order dividing $2$. This means that e.g. the quotient map $\mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ is surjective but not $U_2$-surjective.
If $R$ is Artinian and $M$ is finitely generated, then any injective $R$-linear map $M \rightarrow M$ is surjective. Vasconcelos has also shown that all finitely generated $R$-modules have this property if and only if $R$ has Krull dimension $0.$
You can prove a nice theorem for finite rank free modules: $R^n$ has the property that all injective endomorphisms are surjective if and only if $R$ is a quoring (that is, all regular elements are invertible). Seehere. You have part of the argument already in your question.
Modules $M$ with the property you mention are called co-Hopfian. There is an extensive literature on them, and Lam has some exercises around them. It is worth noting that you can work with $M$ that are not finitely generatd as well. For intance, the abelian group $\oplus \mathbb{Z}(p),$ the sum over the positive primes with $\mathbb{Z}(p)$ the cyclic group of order $p,$ has this property. The property follows since there are no nonzero homomorphism between $\mathbb{Z}(p)$ and $\mathbb{Z}(q)$ if $p \neq q.$
Best Answer
Consider the map $k[x] \to k[x]$ given by multiplication by $x$.
Edit: I feel like I can also shed some light on this duality issue. Do you know what a category is? What we are basically looking at is a category of modules. In category theory there are epimorphisms (epi's) and monomorphisms (mono's) and these are dual to each other: Every category has an opposite category and the epi's in the category correspond to the mono's in the opposite (and vice versa). So for every theorem that can be proven about epi's there is a dual version about mono's (and vice versa).
In module categories the epi's are exactly the surjections and the mono's are exactly the injections. What you've noticed is that in some categories, for instance finitely generated modules over a commutative ring, that epi implies mono. In the dual category it is true that mono implies epi, it's just that the dual category is not the module category for a commutative ring. Hence the dual statement doesn't hold for that category.
You can show that the category of finite dimensional vector spaces is its own dual. So if you prove a statement like "epi implies mono", then the dual statement "mono implies epi" is also true for that category.
So the duality you've noticed between epi's and mono's always holds. It's just it doesn't only involve a single category, it involves looking at a category and it's dual category. What you've noticed is that some categories are their own dual and some aren't.