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 $f(x)=0$.
A few remarks, to be expanded below: (1) first is that the proof that $M = \prod_{i=1}^\infty\mathbb{Z}$ is not free is elementary, and (2) second is that it might be hard to find simpler examples, at least if "simple" refers to how simple the ring is itself.
(1) In fact, here's a proof that I learned from Kaplansky's book "Infinite Abelian Groups": Assume $M$ is free. Pick a prime $p$. Let $S\subseteq M$ be the submodule of sequences such that the power of $p$ dividing each term goes to infinity. $S$ is a submodule of $M$ so is free as well. Multiplication by $(p,p^2,p^3,\dots)$ is an injection $M\hookrightarrow S$. Since $M$ is uncountably generated so is $S$.
Now by definition of $S$, we have that $S/pS$ is a vector space of countable dimension dimension over $\mathbb{Z}/p$ as each term here as a representative with finitely many terms. Yet, if $S$ is free and uncountably generated then $S/pS$ should have uncountable dimension over $\mathbb{Z}/p$.
(2) I have a feeling that finding "easier" examples will be hard. Here are some reasons:
Note that if you did want to try and find an example, you should just experiment with products of the ring $R$ itself. Indeed, S. Chase proved that any product of projectives being projective is equivalent to any arbitrary product of copies of $R$ is projective. (Examples like $\mathbb{Z}/4$ won't work either, since they satisfy the descending chain condition on ideals and all f.g. ideals are finitely related. Any example you are looking for in a ring cannot have these properties.)
Moreover, homologically $\mathbb{Z}$ (and PIDs that are not fields) are pretty simple in that they have global dimension one and they are commutative. Global dimension zero rings won't work since they are semisimple hence quasifrobenius, or equivalently, projectives and injectives coincide. So in this case a product of projectives is a product of injectives, hence injective, hence projective.
I would be interested to see an example however where the ring is more complicated but the proof is even simpler. In fact, I would be interested to see any example at all with a proof significantly different than the one above.
Best Answer
Here is a proof that does not use the axiom of choice. Let $e_n\in A$ denote the sequence whose $n$th term is $1$ and all other terms are $0$; we take it as known that any homomorphism $A\to\mathbb{Z}$ which vanishes on each $e_n$ is $0$ everywhere (the standard proofs of this certainly do not use choice).
Suppose $A$ were projective. Let $F$ be the free group on the underlying set of $A$, and let $p:F\to A$ be the canonical epimorphism. Since $A$ is projective, there is a splitting $i:A\to F$. Now note that for each $a\in A$, there is a homomorphism $\pi_a:F\to\mathbb{Z}$ which takes a formal linear combination of elements of $A$ and gives you the coordinate of $a$. For any $x\in F$, there are only finitely many $a\in A$ such that $\pi_a(x)\neq 0$. In particular, there are only countably many $a\in A$ such that $\pi_a(i(e_n))\neq 0$ for some $n$. But since $\pi_a\circ i:A\to\mathbb{Z}$ vanishes iff it vanishes on each $e_n$, this means that $\pi_a\circ i=0$ for all but countably many $a$. This means that actually the image of $i$ is contained in the free group $G\subset F$ on a countable subset of $A$. But this $G$ is then countable, which is a contradiction since $A$ is uncountable.
(Here I implicitly used the fact that a countable union of finite sets is countable, which requires choice in general. However, the finite sets are finite subsets of $A$, and since it is possible to totally order $A$, we can use this to canonically biject each of the finite subsets with a finite ordinal and thus conclude that their union is countable.)