Prove that this element is nonzero in a tensor product

If you consider any $\mathbb{Z}$-module $M$ and its torsion part $t(M)$, we have the exact sequence $$ 0\to t(M)\to M\to M/t(M)\to0 $$ that, tensored with $\mathbb{Q}$, says $$ \mathbb{Q}\otimes_{\mathbb{Z}}M\cong\mathbb{Q}\otimes_{\mathbb{Z}}M/t(M) $$ So, if $1\otimes x=0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}M$ (for some $x\in M$), then also $1\otimes\pi(x)=0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}M/t(M)$ (for $\pi\colon M\to M/t(M)$ the canonical map).

Now, if $N$ is torsion-free and $y\in N$, $y\ne0$, then $1\otimes y\ne0$ in $\mathbb{Q}\otimes_{\mathbb{Z}}N$, because the map $$ N\to\mathbb{Q}\otimes_{\mathbb{Z}}N,\qquad y\mapsto 1\otimes y $$ is injective.

Note that the first part uses the fact that $\mathbb{Q}$ is divisible (so tensoring with it kills the torsion part); the second part uses that $\mathbb{Q}$ is torsionfree, so flat over $\mathbb{Z}$.

In your case, the element $(1,1,\dotsc)$ has infinite order in $M=\prod_{n>0}\mathbb{Z}/n\mathbb{Z}$, so it goes to a nonzero element in the quotient $M/t(M)$.

Here is an alternative approach which however only works for tensor products that can be considered as localizations: If $R$ is a commutative ring and $S\subset R$ is a multiplicative subset of $R$, then given any $R$-module $M$ the $R_S$-module $R_S\otimes_R M$ together with the map $M\to R_S\otimes_R M$, $m\mapsto 1\otimes m$, is a localization of $M$ at $S$. However, you know that the localization can also be defined as $S^{-1} M$ by taking as elements the equivalence classes of formal fractions $\frac{m}{s}$ with $s\in S$ and $m\in M$ under the relation $\frac{m_0}{s_0} = \frac{m_1}{s_1}:\Leftrightarrow t s_1 m_0 = t s_0 m_1$ for some $t\in S$, together with the map $M\to S^{-1}M$ sending $m$ to $\frac{m}{1}$. By uniqueness of localization, these two approaches are uniquely isomorphic over $R_S$ in a way compatible with the morphisms from $M$ - in particular, given $m\in M$, we have $1\otimes m=0$ in $R_S\otimes_R M$ if and only if there exists some $t\in S$ such that $tm=0$.

This applies to $R := {\mathbb Z}$, $S := {\mathbb Z}\setminus\{0\}$ (so that $R_S = {\mathbb Q}$), proving that the kernel of $M\to M\otimes_{\mathbb Z}{\mathbb Q}$ is precisely the torsion subgroup of $M$. Since in your example the element $(1,1,...)\in\prod\limits_{n\geq 2}{\mathbb Z}/n{\mathbb Z}$ is not torsion, it does therefore not vanish under the localization map to ${\mathbb Q}\otimes_{\mathbb Z}\prod\limits_{n\geq 2}{\mathbb Z}/n{\mathbb Z}$.