Take a field $k$, put $R := k[x]/(x^2)$ and consider the full subcategory ${\mathscr C}$ of $R\text{-Mod}$ consisting of projective (=flat=injective, see below) $R$-modules.
Directed limits of flat modules are always (i.e., for any ring) flat, so ${\mathscr C}$ inherits them from $R\text{-Mod}$.
Products of injectives are always (i.e, in any category) injective, so ${\mathscr C}$ inherits them from $R\text{-Mod}$.
$\alpha: R\xrightarrow{\cdot x} R$ does not have a kernel in ${\mathscr C}$:
First note that since $R\in{\mathscr C}$, a monomorphism in ${\mathscr C}$ is a monomorphism in $R\text{-Mod}$. In particular, a kernel of $\alpha$ in ${\mathscr C}$ would be a subobject of $R$ in $R\text{-Mod}$, of which there are only $\{0\}$, $k\cdot x$ and $R$. $\{0\}$ is excluded since $\alpha^2=0$ but $\alpha\neq 0$, so $\alpha$ is not a monomorphism in ${\mathscr C}$. Similarly, $R$ itself it excluded since $\alpha\neq 0$. Finally, $k\cdot x$ is excluded since it does not belong to ${\mathscr C}$.
Proof that flat, projective and injective modules coincide: Projective modules are flat, any flat module is a directed limit of finitely generated projective modules, and any directed limit of injective modules is injective (since $R$ is Noetherian). Hence, to show that $\text{Proj}\subset\text{Flat}\subset \text{Inj}$ it suffices to note that $R$ itself is injective. For $\text{Inj}\subset\text{Proj}$, take $v\in I\in\text{Inj}$ and suppose $x\cdot v=0$. Then, extending $k\cdot x\hookrightarrow I, x\mapsto v$, along the embedding $k\cdot x\hookrightarrow R$ shows that there exists some $v^{\prime}\in I$ with $x\cdot v^{\prime}=v$. Hence $X := \text{ker}(x\cdot -)=\text{im}(x\cdot -)\subset I$, so for $X^{\prime}$ a $k$-complement of $X$ in $I$, the canonical map $R\otimes_k X^{\prime}\to I$ is an isomorphism, so $I$ is projective.
Note that $\text{Proj}=\text{Inj}=\text{Flat}$ all consist of the "acyclic" $R$-modules, those where $\text{ker}(x\cdot -)=\text{im}(x\cdot -)$.