$\newcommand{\FF}{\mathbb{F}}\DeclareMathOperator{\SL}{SL}$
OK, let me try.

Write $M=E[p]$. If the order of $G$ is coprime to $p$, then $H^1(G,M)=0$. Assume that $p$ divides the order of $G$. Now by Prop 15 in Serre's "Propriétés galoisiennes... " Invent Math 15, $G$ either contains $\SL_2(\FF_p)$ or it is contained in a Borel subgroup. According to the question, we may assume that $G$ is contained in a Borel, say upper triangular matrices.

Let $H=G\cap \SL_2(\FF_p)$. Restriction shows that $H^1(G,M)$ is the $G/H$-fixed part of $H^1(H,M)$. Now $H$ is a subgroup of $(\begin{smallmatrix} a & b \\ 0 & 1/a\end{smallmatrix})$ with invertible $a$ and arbitrary $b$. By assumption $H$ contains the subgroup $K$ generated by $h = (\begin{smallmatrix} 1 & 1\\ 0 & 1\end{smallmatrix})$. Again by restriction-inflation, $H^1(H,M)$ is contained in $H^1(K,M)$. The latter can be computed as usual and we find that a cocyle is determined by its image on $h$ which has to belong to $\FF_p(1, 0)$.

Now we consider again the action of $G/H$ on $H^1(K,M)$. If I am not mistaken, then the class $\bar g$ of matrices of determinant $d$ acts by $(\bar g * \xi)(h) = d \cdot \xi(h)$. If so, then there are no elements in $H^1(H,M)$ fixed by $G/H$ as soon as there is an element of determinant $\neq 1$ in $G$. By the Weil pairing, this must be the case when $p>2$. Hence $H^1(G,M)=0$.

*Edit*: However, I am mistaken as the example of Tyler Lawson shows. The action of $G/H$ may be trivial.