The result follows for arbitrary finite abelian groups from the $p$-group case.
Remember that a finite abelian group $G$ is the direct sum of its $p$-parts,
$$G = G(p_1)\oplus \cdots\oplus G(p_n),$$
where $p_1,\ldots,p_n$ are the distinct primes that divide $|G|$, and
$$G(q) = \{ a\in G\mid q^ma = 0 \text{ for some }m\geq 0\},\qquad q\text{ a prime.}$$
If $a\in G$ is of maximal order, then we can write $a=a_1+a_2+\cdots+a_n$, where $a_i\in G(p_i)$. Since $a$ is of maximal order in $G$, then $a_i$ is of maximal order in $G(p_i)$. By the $p$-group case, we can write $G(p_i) = \langle a_i\rangle\oplus H_i$ with $H_i\leq G(p_i)$. Then $H_1+\cdots+H_n$ is a subgroup of $G$, it is the internal direct sum of the $H_i$, and since $G(p_i) =\langle a_i\rangle\oplus H_i$, then
$$\begin{align*}
G &= G(p_1)\oplus \cdots \oplus G(p_n)\\
&= (\langle a_1\rangle\oplus H_1) \oplus \cdots \oplus (\langle a_n\rangle \oplus H_n)\\
&= (\langle a_1\rangle\oplus\cdots \oplus\langle a_n\rangle) \oplus (H_1\oplus\cdots\oplus H_n).
\end{align*}$$
To finish off, note that $\langle a_1\rangle\oplus\cdots\oplus \langle a_n\rangle = \langle a\rangle$ (e.g., by the Chinese Remainder Theorem).
The tool for this is the Smith Normal Form, a kind of Gaussian elimination for PID.
For the example $G = \langle s,t,u,v \mid s^{4}t^{2}u^{10}v^{6} = s^{8}t^{4}u^{8}v^{10} = s^{6}t^{2}u^{9}v^{8} = e_G \rangle$ written additively, the matrix is
$$
\left(
\begin{array}{cccc}
4 & 2 & 10 & 6 \\
8 & 4 & 8 & 10 \\
6 & 2 & 9 & 8 \\
\end{array}
\right)
$$
whose Smith Normal Form is
$$
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
\end{array}
\right)
$$
as computed by Mathematica using the code mentioned here.
This means that $G$ is generated by four elements $g_1, g_2, g_3, g_4$ such that $1g_1=0$, $2g_2=0$, $2g_3=0$, and no restrictions on $g_4$, and so $G \cong C_1 \times C_2 \times C_2 \times C_{\infty} \cong C_2 \times C_2 \times C_{\infty}$.
$g_1, g_2, g_3, g_4$ are obtained from $s,t,u,v$ by applying the matrices $P$ and $Q$ such that $PAQ$ is the diagonal matrix above.
All this is explained in several books. One is
Finitely Generated Abelian Groups and Similarity of Matrices over a Field by Christopher Norman. A shorter account appears in Jacobson's Basic Algebra I.
Best Answer
As pointed out by i707107, the problem is essnetially equvalent to find the Smith normal form of the $3\times 3$ matrix which is defined by the relations. One can find the details of the algorithm to find the Smith normal form in Wikipedia as linked above. I just do the calculation and find the group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.