Let $F$ be a field. Then $SL(n,F)$ is the set of all $n\times n$ matrices with determinant $1$ (equivalently, the sum of the main diagonal is $0$). We can easily show that $SL(n,F)$ is a vector space under matrix addition and multiplication by scalars from $F$:
$i)$ $\hat 0 \in SL(n,F)$, as $\displaystyle{\sum_{i=1}^n0=0}$.
$ii)$ Let $A$, $B$ $\in SL(n,F)$. Then $\displaystyle{\sum_{i=1}^n a_{ii}=0}$ and $\displaystyle{\sum_{i=1}^n b_{ii}=0}$. Adding these together, $\displaystyle{\sum_{i=1}^n a_{ii}+\displaystyle{\sum_{i=1}^n b_{ii}=\displaystyle{\sum_{i=1}^n a_{ii}+b_{ii}=0}}}$. Thus $A+B\in SL(n,F)$.
$iii)$ Let $A\in SL(n,F)$, $\alpha \in F$. Consider $\alpha A$: $\alpha\displaystyle{\sum_{i=1}^n a_{ii}=\displaystyle{\sum_{i=1}^n \alpha a_{ii}=0}}$. Thus $\alpha A \in SL(n,F)$.
Therefore $SL(n,F)$ is a vector space. What is a basis for $SL(n,F)$?
The matrices with determinant $1$ are not a subspace of $\def\M{{\rm Mat}}\M_n(F)$: For we have $$ \det(a\def\I{\mathrm{Id}}\I) = a^n \det \I = a^n $$ so $a \I \in \def\SL{\mathrm{SL}}\SL_n(F)$ iff $a^n = 1$, which does not hold for all $a \in F$, espacially not for $a = 0$. (Note that $\det 0 = 0 \ne 1$, hence $0 \not\in \SL_n(F)$.
The matrices with trace 0 on the other hand (which form the Lie algebra $\def\sl{\mathfrak{sl}}\sl_n(F)$ associated with the group $\SL_n(F)$) are a vector subspace of $\M_n(F)$ -- as you showed above. A basis is given as follows: Let $E_{i,j}$ denote the matrix which has only zeros but a one at the $(i,j)$ spot as entries. Then $(E_{i,j} \mid i, j = 1, \ldots, n)$ is a basis of $\M_n(F)$ and $(E_{i,j} \mid i \ne j) \cup (E_{i,i} - E_{1,1} \mid i = 2,\ldots n)$ is a basis of the $n^2 - 1$-dimensional space $\sl_n(F)$.