$\begin{array}{ccccccccc} \times&e_0&e_1&e_2&e_3&e_4&e_5&e_6&e_7\\ e_0&I&e_1&e_2&e_3&e_4&e_5&e_6&e_7\\ e_1&e_1&-I&e_3&-e_2&e_5&-e_4&e_7&-e_6\\ e_2&e_2&e_3&-I&-e_1&e_6&e_7&-e_4&-e_5\\ e_3&e_3&-e_2&-e_1&I&e_7&-e_6&-e_5&e_4\\ e_4&e_4&e_5&-e_6&-e_7&-I&-e_1&e_2&e_3\\ e_5&e_5&-e_4&-e_7&e_6&-e_1&I&e_3&-e_2\\ e_6&e_6&e_7&e_4&e_5&-e_2&-e_3&-I&-e_1\\ e_7&e_7&-e_6&e_5&-e_4&-e_3&e_2&-e_1&I \end{array}$
There is an identity element and each element has an inverse but is the multiplication associative? To check manually would require $2\times8^3$ multiplications. I'm hoping there is a more elegant way than by brute force. If this is a group, is it $Q_8$?
If it helps, $e_0,e_2,e_4,e_6$ are quaternions.
It is the group $(C_4\times Q_8)/\{(0,0),(2,-1)\}$ with $$ \begin{array}{c} e_0 = (0,1) \qquad e_2 = (0,i) \qquad e_4 = (0,j) \qquad e_6 = (0,k) \\ e_1 = (1,1) \qquad e_3 = (1,i) \qquad e_5 = (1,j) \qquad e_7 = (1,k) \end{array}$$
This can also be presented as the group generated by $\{e_1,e_2,e_4\}$ with the relations $$ e_1^4=1 \qquad e_1^2=e_2^2=e_4^2 \qquad e_1e_2=e_2e_1 \qquad e_1e_4=e_4e_1 \qquad e_4e_2 = e_1^2e_2e_4 $$
Then, in your table, $e_0=1$ and $ e_6=e_2e_4 $ and $ e_{2n+1} = e_1e_{2n} $ and $-e_n = e_2^2e_n $.