Can a torsion group and a nontorsion group be elementarily equivalent? I think that the direct product
$\prod_{n \in {\bf N}\setminus \{0\}} {\bf Z}/(n)$
and the direct sum
$\bigoplus_{n \in {\bf N} \setminus \{0\}} {\bf Z}/(n)$
are elementarily equivalent but am not sure how to prove it. 
 A: The question in the title is an easy application of compactness. Let $G$ be a torsion group containing elements of arbitrarily large order (your group $\bigoplus_{n\in \mathbb{N}_+}\mathbb{Z}/(n)$ will do nicely). Let $c$ be a new constant symbol, and consider the theory $T' = \text{Th}(G)\cup \{n\cdot c \neq 0\mid n\in \mathbb{N}_+\}$ (where $n\cdot c$ is an abbreviation for the sum of $n$ copies of $c$). By compactness, this theory has a model, which is a non-torsion group elementarily equivalent to $G$ (given a finite subset of $T'$, take $G$ and interpret $c$ as an element of large enough order). [I've just seen Noah's answer - note the parallels between the compactness argument and the ultrapower argument. This is a very general pattern.]
But the problem posed in the body is more complicated. I suspect that the direct sum is an elementary substructure of the direct product. You might be able to show this using the Tarski-Vaught test, together with the quantifier elimination down to positive primitive formulas for abelian groups (see Theorem 3.3.5 in A Course in Model Theory by Tent and Ziegler for the general case of $R$-modules, and take $R = \mathbb{Z}$ to specialize to abelian groups).
A: Yes, a torsion group can indeed be elementarily equivalent to a non-torsion group. 
We can get a (somewhat) explicit example using the following fact, together with Łoś's theorem:

If $G$ is torsion, but has elements of arbitrarily large finite order, then any nontrivial ultrapower over $\mathbb{N}$ of $G$ is non-torsion.

Proof: suppose $G$ is as above, and $\mathcal{U}$ is a nonprincipal ultrafilter on $\mathbb{N}$. Then let $a_i\in G$ have order $>i$, and consider the element $$\alpha=[(a_i)_{i\in\mathbb{N}}]_\mathcal{U}$$ of the ultrapower $\prod_{\mathbb{N}}G/\mathcal{U}$. For each $i$, the formula "$x$ has order $>i$" holds of cofinitely many $a_j$s (namely, each $a_j$ for $j\ge i$); since $\mathcal{U}$ is nonprincipal, every cofinite set is in $\mathcal{U}$, so $\alpha$ has infinite order.
As usual, this ultrapower argument can be replaced by a simple compactness argument - see Alex Kruckman's answer. I gave the ultrapower construction above since I think it's cool.

I believe that in particular, the groups you mention in your question are elementarily equivalent, but I don't immediately see how to show that (I suspect a proof via Ehrenfeucht-Fraïssé games wouldn't be too hard though).
