This seems too easy, so it is likely not correct, but would be interested to know why.
Let $(M\oplus N\to Q_i):={\rm coker} (P_i\to M\oplus N)$ so one has short exact sequences: $$ 0\to P_i \to M\oplus N \to Q_i \to 0. $$ Then (I think, and this could be a sticky point) $$ {\rm lim} P_i = M\oplus N \quad \Leftrightarrow \quad {\rm lim}\, Q_i =0. $$ Now let $(K_i\to M):={\rm ker}(M\to M\oplus N\to Q_i)$. Since ${\rm lim}\\, Q_i =0$, ${\rm lim}K_i=M$ and hence there exists an $i$ such that $K_i=M$ and for that $i$ (and all that follows) $M\to Q_i$ is the zero map. Similarly for $N$, so we have an $i$ such that both $M\to Q_i$ and $N\to Q_i$ are zero, so $M\oplus N\to Q_i$ is also zero, but then $P_i={\rm ker}(M\oplus N\to Q_i)=M\oplus N$.