Let $G=<x> \times <y>$ where $|x|=|y|=p$ and so $|G|=p^2$. Let $k$ be a field of characteristic $p$. Let $W$ be the $k$-span of $v$ and $u$. We wish to show the module structure given by
$$x \cdot u = u + \alpha v$$
$$x \cdot v = v$$
$$y \cdot u = u + \beta v$$
$$y \cdot v = v$$
where $\alpha,\beta \in k$.
We can represent $x$ and $y$ by matrices
$$x = \begin{pmatrix} 1 & 0\\ \alpha & 1 \end{pmatrix}$$
$$y = \begin{pmatrix} 1 & 0\\ \beta & 1 \end{pmatrix}$$
If I understand: To show this is structure is unique we are essentially be asked to show that the only commutative set of $2 \times 2$ matrices which commute and are of order $p$ are of the form given above. However, this seems to miss a lot of what I might already know about representation theory and so I wanted to confirm that really is the approach and how to continue from that idea if that is the case.
For a $k$ algebra $A$, to show that "$M$ has a unique left $A$ module structure" amounts to showing there is exactly one algebra homomorphism $A\to End(M_k)$.
To show that it has "a unique module structure such that..." you'd be showing there is one homomorphism with those properties.
Now, we're assisted in the fact that there is a universal property that if $A$ is an $R$ algebra, then any group homomorphism $G\to U(A)$ extends uniquely to an algebra homomorphism $R[G]\to A$.
In your case, $R=k$ and $A=M_2(k)$, and the assignments you provided make a group homomorphism $G\to U(M_2(k))= GL_2(k)$. Therefore by the universal property this extends uniquely to an algebra homomorphism $k[G]\to M_2(k)$.