The following exercise is taken from T.S. Blyth book, Module Theory, chapter 6, exercise 9:
Something does not add up, especially with using (1). It seems that condition (1) is not necessary or sufficient here and the statement can be proved solely by (2).
This is because the $R$-morphism $h:M\rightarrow\prod_{i\in I}M_i$ which can be defined by $h(x):=(\pi_i(x))_{i\in I}$ is exactly an Isomorphism if and only if (2) holds. And $h$ is an Isomorphism if and only if $(M,(\pi_i(x))_{i\in I})$ is a product of the family $(M_i)_{i\in I}$. Am I missing something? I checked and re-checked over and over and I don't see why I need (1).
Definition: Here is definition of category.
Let $\mathcal{C}$ be a category and $\{X_j\mid j\in I\}$ be family of object of $\mathcal{C}$. A product for the family $\{X_j\mid j\in I\}$ is an object $X$ of $\mathcal{C}$ together with a family of morphisms $\{ \pi_j : X\to X_j \mid j\in I\}$ such that for any object $Y$ and family of morphisms $\{\varphi_j:Y\to X_j\mid j\in I\}$, there is a unique morphism $\varphi : Y\to X$ such that $\pi_j\circ \varphi =\varphi_j$ for all $j\in I$.
Product of family $\{X_j\mid j\in I\}$ is unique, in a sense, if $(P,\{\pi_j\})$ and $(Q,\{\psi_j\})$ are both products of the family $\{X_j\mid j\in I\}$ of objects of a category $\mathcal{C}$, then $P$ and $Q$ are equivalent.
Proof: Let $\mathcal{C}$ be category of $R$-modules. $(\Rightarrow)$ Suppose $(M,\{\pi_j\})$ is product of family $\{M_j\mid j\in I\}$. Let $\prod_{j\in I}M_j$ be cartesian product of possibly infinite family $\{M_j\mid j\in I\}$, and let $\pi_j’: \prod M_j \to M_j$ be canonical projection for all $j\in I$. Then $(\prod M_j, \{\pi_j’\})$ is product of family $\{M_j\mid j\in I\}$ is quite a standard result. So $M$ and $\prod M_j$ are equivalent ($R$-module isomorphic).
Let $M\cong \prod M_j$ under an isomorphism $f:M\to \prod M_j$. Then $\pi_j=\pi_j’f$ by the definition of product. Let $i_j’:M_j\to \prod M_j$ be canonical injection. For all $j\in I$, define $i_j:=f^{-1}i_j’:M_j\to M$. So $$\pi_k\circ i_j=\pi_k’f\circ f^{-1}i_j’=\pi_k’\circ i_j’= \begin{cases} \text{id}_{M_j} & \text{if $k=j$} \\ 0 & \text{if $k\neq j$}\end{cases}.$$ Condition $(1)$ satisfy. Since $f$ is equivalence (bijective), for every $(x_j)_{j\in I}\in \prod M_j$ there exists unique $x\in M$ such that $f(x)=(x_j)_{j\in I}$. So $$\pi_j(x)=\pi_j’\circ f(x)=\pi_j’((x_j)_{j\in I})=x_j.$$ Condition $(2)$ satisfy. Thus $\Rightarrow$ holds.
$(\Leftarrow)$ Suppose family $\{i_j:M_j\to M\mid j\in I\}$ of morphism ($R$-module homomorphism) satisfies condition $(1)$ and $(2)$. We want to show $(M,\{\pi_j\})$ is product of family $\{M_j\mid j\in I\}$. Let $A$ be an object in $\mathcal{C}$ and family of morphism $\{\varphi_j:A\to M_j\mid j\in I\}$. Let $a\in A$. Then $(\varphi_j(a))_{j\in I}\in \prod M_j$. By condition $(2)$, $\exists !x_a\in M$ such that $\pi_j(x_a)=\varphi_j (a)$ for all $j\in I$.
Define $\varphi : A\to M$ given by $\varphi (a)=x_a$. Note $\varphi$ is a well defined function. So $$\pi_j\circ \varphi (a)=\pi_j (x_a)=\varphi_j(a).$$ Thus $\pi_j\circ \varphi (a)= \varphi_j(a)$ for all $a\in A$. Therefore $\pi_j\circ \varphi = \varphi_j$ for all $j\in J$. Let $\psi :A\to M$ such that $\pi_j\circ \psi =\varphi_j$ for all $j\in I$. Fix $a\in A$. Then $\pi_j(\psi (a))=\varphi_j (a)\in M_j$. By uniqueness of $x_a$, we have $\psi (a)=x_a=\varphi (a)$. So $\varphi = \psi$. Thus $\exists !\varphi :A\to M$ such that $\pi_j\circ \varphi =\varphi_j$ for all $j\in I$. Hence $(M,\{\pi_j\})$ is product of $\{M_j\mid j\in I\}$. In proof of $\Leftarrow$ we didn’t make use of condition $(1)$.
TL;DR You were correct in your assessment of condition $(1)$ is not needed in the exercise.