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.