Retraction and direct limit

No, not necessarily. For instance, let $A=\mathbb{Z}$. Since $\mathbb{Q}$ is not projective as an $A$-module, there is some short exact sequence $$0\to K\to M\stackrel{p}{\to}\mathbb{Q}\to 0$$ that does not split (explicitly, you could take $M$ to be any free module with a surjection to $\mathbb{Q}$). Now let $M_i=p^{-1}(\frac{1}{i!}\mathbb{Z})\subset M$. Then for each $i$, our short exact sequence restricts to a short exact sequence $$0\to K\to M_i\to \frac{1}{i!}\mathbb{Z}\to 0$$ which splits since $\frac{1}{i!}\mathbb{Z}$ is free. This means that the inclusion $K\to M_i$ has a retraction for all $i$, but the colimit $K\to M$ does not.

More generally, you can get a similar example from any non-projective module that is the union of an ascending sequence of projective submodules.

If there was a retraction $s:M\rightarrow M_0$, composition with the canonical morphisms $\nu_i: M_i \rightarrow M$ gives a system of retractions $s_i:M_i \rightarrow M_0$ that is compatible with the direct system in the sense that $s_{i+1}\iota_i=s_i,$ where $\iota_i: M_i \rightarrow M_{i+1}$ are the inclusions of the direct system. Thus, to disprove the claim it is enough to find a direct system that allows for retractions while does not allow for a system of compatible retractions.

Let us try to cook up some example of this where $M_i$ is the free Abelian group of rank $2$ for all $i >0$, and $M_0=\mathbb{Z}$. By choosing coordinates on $M_i$'s suitably, we may assume that the inlusions $M_0 \subseteq M_i$ coming from the direct system are of the form $$\begin{pmatrix}0 \\ 1\end{pmatrix}: \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z},\;\; k \mapsto \begin{pmatrix}0 \\ k\end{pmatrix}.$$

Then a retraction for any $M_i \rightarrow M_0$ is necessarily of the form $$\begin{pmatrix}k_i & 1\end{pmatrix}: \mathbb{Z}\oplus \mathbb{Z} \rightarrow \mathbb{Z},$$ the inclusions $\iota_i: M_i \rightarrow M_{i+1}$ are necessarily of the form $$\begin{pmatrix}m_i & 0 \\ n_i & 1\end{pmatrix}:\mathbb{Z}\oplus \mathbb{Z} \rightarrow \mathbb{Z}\oplus \mathbb{Z}$$ (this form is forced upon us by the condition $A \cdot\begin{pmatrix}0 \\ 1\end{pmatrix}=\begin{pmatrix}0 \\ 1\end{pmatrix}$), and we can also easily check that such a map is injective whenever $m_i \neq 0$.

So far, the situation was completely general, i.e. we did not make any real choices (we just considered suitably chosen coordinates). Let us now set $m_i=n_i=3$ for all $i>0$, i.e. consider a system where all the maps $\iota_i$ are, with the coordinate choice we made, of the form $$\begin{pmatrix}3 & 0 \\ 3 & 1\end{pmatrix}:\mathbb{Z}\oplus \mathbb{Z} \rightarrow \mathbb{Z}\oplus \mathbb{Z}, \;\; \begin{pmatrix}a \\ b\end{pmatrix} \mapsto \begin{pmatrix}3a\\ 3a +b\end{pmatrix}.$$ Then to have a compatible system of retractions $$M_i \rightarrow M_0$$ is to have $k_i$ satisfying $$k_{i}=3k_{i+1}+3,\; \text{ i.e. }\;k_{i+1}=\frac{k_i}{3}-1$$

Now we can readily check that such collection of $k_i's$ cannot exist: The last recurrence relation allows for explicit form $$k_{i+1}=\left(\frac{1}{3}\right)^i(k_1+\frac{3}{2})-\frac{3}{2},$$ and one can easily check that regardless of $k_1 \in \mathbb{Z}$, $k_{i+1}$ is not an integer for sufficiently large $i$.

Altogether, such system does not allow for a compatible system of retractions, hence there is no retraction to $M_0$ from the direct limit.