If $A$ is an abelian group, then $\cap_{n \geq 0} 2^n A$ is a subgroup of $A$, whose elements may be called $2^{\infty}$-divisible. Note that $0$ is the only $2^{\infty}$-divisible element of $\mathbb{Z}$. Therefore, the same is true for $\mathbb{Z}^{\mathbb{N}}$. But the element represented by $(2^0,2^1,2^2,\dotsc)$ is $2^{\infty}$-divisible in $\mathbb{Z}^{\mathbb{N}}/\mathbb{Z}^{\oplus \mathbb{N}}$. Hence, there is no monomorphism $\mathbb{Z}^{\mathbb{N}}/\mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$.
Alternatively, it is a well-known result by Baer that $\mathbb{Z}^{\oplus \mathbb{N}} \to \hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$, $e_n \mapsto \mathrm{pr_n}$ is an isomorphism. In particular (and actually this is the main step in the proof) a homomorphism $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}$ vanishes when it does vanish on the direct sum, i.e. $\hom(\mathbb{Z}^{\mathbb{N}}/\mathbb{Z}^{\oplus \mathbb{N}},\mathbb{Z})=0$.
For the sake of completeness, here is the argument: If $f : \mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}$ is a homomorphism which vanishes on the direct sum, and $x \in \mathbb{Z}^\mathbb{N}$, then for every $n \in \mathbb{N}$ we choose $u_n,v_n \in \mathbb{Z}$ with $x_n = 2^n \cdot u_n + 3^n \cdot v_n$. Then one observes that $f(2^n u_n)_n \in \mathbb{Z}$ is $2^{\infty}$-divisible, thus vanishes. Likewise $f(3^n v_n)_n$ vanishes, so that $f(x)=0$.
This answer here is solely for the purpose of giving this question an answer. Since the OP obtained the desired answer (see the comments under the question), I am providing a different way using category theory to show that $$P:=\frac{\prod\limits_{\alpha\in J}\,A_\alpha}{\prod\limits_{\alpha\in J}\,B_\alpha}\cong \prod_{\alpha\in J}\,\frac{A_\alpha}{B_\alpha}\text{ and }S:=\frac{\bigoplus\limits_{\alpha\in J}\,A_\alpha}{\bigoplus\limits_{\alpha\in J}\,B_\alpha}\cong \bigoplus_{\alpha\in J}\,\frac{A_\alpha}{B_\alpha}\,.$$
Explicit isomorphisms can be seen in (*) and (#).
For each $\beta \in J$, $\iota_\beta:A_\beta\to \bigoplus\limits_{\alpha\in J}\,A_\alpha$ and $\pi_\beta: \prod\limits_{\alpha\in J}\,A_\alpha\to A_\beta$ denote the canonical injection and the canonical projection, respectively. Let $q:\bigoplus\limits_{\alpha\in J}\,A_\alpha\to S$ be the quotient map. Then, $q\circ \iota_\beta$ vanishes on $B_\beta$. Therefore, $q\circ \iota_\beta$ factors through the quotient map $q_\beta:A_\beta\to\dfrac{A_\beta}{B_\beta}$. In other words, there exists a (unique) map $i_\beta:\dfrac{A_\beta}{B_\beta}\to S$ such that $$q\circ \iota_\beta=i_\beta\circ q_\beta\,.$$
We claim that $S$ together with the maps $i_\beta:\dfrac{A_\beta}{B_\beta}\to S$ for $\beta\in J$ is a categorical coproduct (direct sum) of the family $\left(\dfrac{A_\alpha}{B_\alpha}\right)_{\alpha\in J}$. Let $T$ be any $R$-module together with morphisms $\tau_\beta:\dfrac{A_\beta}{B_\beta}\to T$ for each $\beta\in J$. We want to show that a there exists a unique morphism $\phi:S\to T$ such that $\phi\circ i_\beta=\tau_\beta$ for each $\beta\in J$.
We define
$$\phi\left((a_\alpha)_{\alpha\in J}+\bigoplus_{\alpha\in J}\,B_\alpha\right):=\sum_{\alpha\in J}\,\tau_\alpha\left(a_\alpha+B_\alpha\right)\text{ for all }(a_\alpha)_{\alpha\in J}\in\bigoplus_{\alpha\in J}\,A_\alpha\,.$$
It is easy to verified that $\phi$ is a well defined morphism, and it is the only morphism such that $\phi\circ i_\beta=\tau_\beta$ for all $\beta\in J$. We can now then conclude that $S$ is a coproduct of the family $\left(\dfrac{A_\alpha}{B_\alpha}\right)_{\alpha\in J}$. Since coproducts are unique up to isomorphism, we obtain $S\cong \bigoplus\limits_{\alpha \in J}\,\dfrac{A_\alpha}{B_\alpha}$, via the isomorphism $\sigma:S\to \bigoplus\limits_{\alpha \in J}\,\dfrac{A_\alpha}{B_\alpha}$ given by
$$\sigma\left((a_\alpha)_{\alpha\in J}+\bigoplus_{\alpha\in J}\,B_\alpha\right):=\sum_{\alpha\in J}\,\bar{\iota}_\alpha\left(a_\alpha+B_\alpha\right)\text{ for all }(a_\alpha)_{\alpha\in J}\in\bigoplus_{\alpha\in J}\,A_\alpha\,,\tag{*}$$
where $\bar{\iota}_\beta:\dfrac{A_\beta}{B_\beta}\to \bigoplus\limits_{\alpha\in J}\,\dfrac{A_\alpha}{B_\alpha}$ is the canonical injection for each $\beta\in J$.
Observe now that, for every $\beta\in J$, $q_\beta\circ \pi_\beta$ vanishes on $\prod\limits_{\alpha\in J}\,B_\alpha$. Therefore, $q_\beta\circ\pi_\beta$ factors through the quotient map $k:\prod\limits_{\alpha\in J}\,A_\alpha\to P$. Ergo, there exists a (unique) morphism $\varpi_\beta:P\to \dfrac{A_\beta}{B_\beta}$ such that $$q_\beta\circ\pi_\beta=\varpi_\beta\circ k\,.$$ We claim that $P$ together with the morphisms $\varpi:P\to \dfrac{A_\beta}{B_\beta}$ is a categorical product (direct product) of the family $\left(\dfrac{A_\alpha}{B_\alpha}\right)_{\alpha\in J}$. Let $Q$ be any $R$-module together with morphisms $\kappa_\beta:Q\to\dfrac{A_\beta}{B_\beta}$ for all $\beta\in J$. We need to show that there exists a unique morphism $\psi:Q\to P$ such that $\varpi_\beta\circ \psi=\kappa_\beta$ for all $\beta\in J$.
We define
$$\psi\left(x\right):=\big(\kappa_\alpha(x)\big)_{\alpha\in J}+\prod_{\alpha\in J}\,B_\alpha\text{ for all }x\in Q\,.$$
It is easily seen that $\psi$ is a well defined morphism, and it is the only morphism such that $\varpi_\beta\circ \psi=\kappa_\beta$ for all $\beta\in J$. We now conclude that $P$ is indeed a product of the family $\left(\dfrac{A_\alpha}{B_\alpha}\right)_{\alpha\in J}$. Since products are unique up to isomorphism, we have $P\cong \prod\limits_{\alpha\in J}\,\dfrac{A_\alpha}{B_\alpha}$ via the isomorphism $\varsigma: \prod\limits_{\alpha\in J}\,\dfrac{A_\alpha}{B_\alpha}\to P$ given by
$$\varsigma\Big(\big(a_\alpha+B_\beta\big)_{\alpha\in J}\Big):=\big(a_\alpha\big)_{\alpha\in J}+\prod_{\alpha\in J}\,B_\alpha\text{ for all }\big(a_\alpha\big)_{\alpha\in J}\in \prod_{\alpha\in J}\,A_\alpha\,.\tag{#}$$
Best Answer
tl;dr Infinite products/sums are commutative and associative as well. Therefore your proof can be easily generalized.
Long answer: Ok, so infinite products/direct sums are more problematic then finite as you will see soon. First of all: what is an infinite product? The definition goes as follows:
If each $A_i$ is a module then this generalized product has a structure of a module as well via pointwise addition and ring action. Under the Axiom of Choice of course (which guarantees that $\prod A_i$ is nonempty). Then the direct sum $\bigoplus_{i\in I}A_i$ is a special subset of $\prod A_i$ consisting of all choice functions which are constant $0$ almost everywhere.
This is important because only by looking at the definition you can have your first "oh" moment: the definition does not depend on any particular order. In other words $A\times B$ is literally equal to $B\times A$ under this definition. And the ordering of such product boils down to ordering of the underlying set of indices, which is not mandatory. It is then a theorem that the classical Cartesian product is isomorphic to this generalized product in the case of finitely many components. Which also shows that the order in the classical Cartesian product is indeed irrelevant in this particular case. This solves your "commutativity" issue.
The other thing you need in this setup is the following variant of the associativity:
Proof. I will use the classical "pair" definition for the outer Cartesian product "$\times$" (just because nesting choice functions is a pain in the ***). With that define $$\Psi:\prod_{i\in I}A_i\to\bigg(\prod_{j\in J}A_j\bigg)\times\bigg(\prod_{k\in K}A_k\bigg)$$ $$\Psi(f)=\left(f_{|J},\ f_{|K}\right)$$ where $f_{|J}$ denotes the classical function restriction. I leave as an exercise that $\Psi$ is an isomorphism. $\Box$
And the last component is:
which you prove similarly.
Finally I leave as an exercise that both lemmas hold for the direct sum as well. All those facts together give you your generalized version of the direct sum of projective modules.
Note that you need to use "direct sums" instead of "products" because in general the (infinite) product of free modules need not be free, even over such nice rings like $\mathbb{Z}$. While the direct sum of free modules is always free.