You are right in that you may have many possible bases for a vector space. But, the ones that you have listed are not bases for the vector space that you are considering. Note that the space of $2\times 2$ matrices (over say the real numbers) is the set of all matrices of the form $ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $ where $a,b,c,$ and $d$ each can be any real number. Now, to see that your first option doesn't work as a basis, you could consider the matrix
$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $ The If the first option was a basis, then you would have to be able to find a real number $s$ such that $ s\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $ But here the left hand side is equal to $ \begin{pmatrix} s & s \\ s & s \end{pmatrix}. $ Noting that two matrices are equal if and only if each entry is equal, that would mean that $s$ would have to be $0$ and $1$, and there are no solutions for this. So you can say that the element $\left(\begin{smallmatrix}1 & 1 \\ 1 & 1 \end{smallmatrix}\right)$ does not span the space of $2\times 2$ matrices.
Now you could try to think about your other option in the same manner. Maybe this would lead you to a correct basis. As others have already noted, a correct basis would have $4$ elements.
As a sidenote and just to make sure that you understand. The span of the elements in, for example, your second option would be $ s\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} + t\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} s & t \\ s & t \end{pmatrix}. $ But again, this does not span the space of $2\times 2$ matrices.