Let $R$ be a commutative ring with unity. For any $x\in R$, let the endomorphism of $M$ induced by multiplication by $x$ be $f_x$, ie., $f_x(m)=mx$. Prove the following:
(1) If $f_x$ is surjective whenever it is injective on $M$, then $f_x$ is surjective whenever it is injective on $\bigoplus_{i\in \Bbb N}M$.
(2) Does it follow that if $f_x$ is injective whenever it is surjective on $M$, then also $f_x$ is injective whenever it is surjective on $\bigoplus_{i\in \Bbb N}M$.
For (2) I have no idea.
For (1), the response is in the paper of KAMRAN DIVAANI-AAZAR AND AMIR MAFI (2004) ``A NEW CHARACTERIZATION OF COMMUTATIVE ARTINIAN RINGS'' in Example 2.2. Modules satisfying (1) are referred to as semi co-Hopfian in this paper. Unfortunately, there is no clear elaboration on the assertion.
Call the statement $f_x$ is injective if and only if it is surjective on $N$ "$f_x$ has $P$ on $N$". I think the following works: $$ f_x\:\text{has $P$ on $\bigoplus_{i\in I}M$}\iff f_x\:\text{has $P$ on $M$ for all $i\in I$}$$ $$ \iff f_x\:\text{has $P$ on $M$}.$$ Indeed, $f_x$ acts on an element $(m_i)_{i\in I}\in \bigoplus_{i\in I} M$ by $f_x(m_i)_{i\in I}=(xm_i)_{i\in I}$, which is just the action of $f_x$ on each $M$. Since this map sends the $i^{th}$ component of the direct sum into itself, it is surjective if and only if it is surjective on each component and injective if and only if it is injective on each component.