Define the annihilator of $a\in R$ with respect the $R$-module $M$ as $(0:_Ma)=\{m\in M: ma=0\}$.
Let $M_1$ and $M_2$ be $R$-modules and $a_1,a_2\in R$, I have been able to show that $$((0,0):_{M_1\times M_2}(a_{1},a_{2}))=(0:_{M_1}a_{1})\times(0:_{M_2}a_{2}).$$ From this, I can extend the proof to the module $\prod_{i=1}^nM_i$ by showing that for any $a_1,a_2,\ldots,a_n\in R$, $$((0,0,\ldots, 0):_{\prod_{i=1}^nM_i}(a_{1},a_{2},\ldots,a_{n}))=(0:_{M_1}a_{1})\times(0:_{M_2}a_{2})\cdots(0:_{M_n}a_{n}).$$
Now, I have failed to extend the same expansion to this direct product, $\prod_{\lambda\in\Lambda}M_\lambda$ where $M_\lambda$ is an $R$-module.
Assist me, please.
There is no need to modify your argument, since the statement follows directly from definitions.
Indeed, let $M = \Pi_\lambda M_\lambda$. Then $$ (0:_M a) = \lbrace (x_\lambda )\in M\,\vert\, (x_\lambda)a = 0\rbrace. $$ But $(x_\lambda )a = (x_\lambda a)=0$ iff $x_\lambda a= 0$ for each $\lambda$. (By definition, $0 = (0_\lambda )$ in $M$. Also, equality of tuples is checked coordinatewise.) In other words, we just have $$ (0:_M a) = \Pi_\lambda \lbrace x\in M_\lambda\,\vert\,x a= 0\rbrace=\Pi_\lambda(0:_{M_\lambda}a). $$