First, $\mathbb{Z}_n \oplus \mathbb{Z}_n$ is abelian, while there are many non-cyclic groups that are non-abelian (take $S_3$ for example), so the answer to your question as written is immediately no.
However, what if we only consider abelian non-cyclic groups? Then $\mathbb{Z}_2 \oplus \mathbb{Z}_6$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ are two counterexamples you might consider. [After OP's edit: a counterexample is $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$, which has order $16$ but is not $\mathbb{Z}_4 \oplus \mathbb{Z}_4$.]
[Note: this is a relevant result that you might already know: if $m$ and $n$ are coprime, then $\mathbb{Z}_m \oplus \mathbb{Z}_n \cong \mathbb{Z}_{mn}$.]
What you might then ask is if every abelian group can be written as the direct product of cyclic groups, and this is true, but not obvious: Classification of finitely generated abelian groups.
Your map $\Phi$ isn't to $GL_2(\Bbb R)$ itself, but rather the subgroup $G = \left\{\pmatrix{a & b \\ -b & a}: a, b \in \Bbb R\right\}$. The instructions aren't incredibly explicit, but it is very important here that $\Phi$ has codomain $G$, rather than all of $GL_2(\Bbb R)$.
When you pick a matrix $M =\pmatrix{m&n\\p&q}$, you're not necessarily picking something in $G$, the codomain of $\Phi$. It may very well be the case that there's no $z \in \Bbb C^*$ such that $\Phi(z) = M$; this happens if and only if $q = m$ and $p = -n$. So if you pick the $M$ above as your generic matrix and try to find some $z \in \Bbb C^*$ so that $\Phi(z) = M$, you had better not be able to show that $\Phi$ is onto $GL_2(\Bbb R)$ -- it's not!
Instead, to show that $\Phi$ is surjective ("onto"), pick something in the codomain of $\Phi$, something in $G$.
So pick a generic matrix $M = \pmatrix{m&n\\-n&m} \in G$, and find the corresponding $z \in \Bbb C^*$ with $\Phi(z) = M$.
In actuality, defining $\Phi$ as above but making its codomain all of $GL_2(\Bbb R)$ instead of just $G$ isn't a huge problem. It's a general fact that for any group homomorphism $\varphi: H \to H'$, if $\varphi$ is injective, then $H$ is isomorphic to the image $\operatorname{im}(\varphi) = \{\varphi(h) : h \in H\}$ (this is a consequence of the first isomorphism theorem).
Since you've shown that $\Phi$ is an injective homomorphism from $\Bbb C^*$, then $\Bbb C^*$ is necessarily isomorphic to $\operatorname{im}(\Phi)$, which is exactly the $G$ defined in the problem statement. It's just that, unless the codomain of $\Phi$ is exactly $\operatorname{im}(\Phi)$, you won't be able to show that $\Phi$ is surjective.
Best Answer
Consider the group of those matrices of the form\begin{pmatrix}\cos\left(\frac{2k\pi}n\right)&-\sin\left(\frac{2k\pi}n\right)\\\sin\left(\frac{2k\pi}n\right)&\cos\left(\frac{2k\pi}n\right)\end{pmatrix}($k\in\{0,1,\ldots,n-1\}$). In other words, it is the subgroup of $GL_2(\mathbb{R})$ generated by the matrix$$\begin{pmatrix}\cos\left(\frac{2\pi}n\right)&-\sin\left(\frac{2\pi}n\right)\\\sin\left(\frac{2\pi}n\right)&\cos\left(\frac{2\pi}n\right)\end{pmatrix},$$which has order $n$.