The following is my proof:
Theorem: If $W$ is the space of $n \times n$ matrices over the field $F$ and if $f$ is a linear functional on $W$ such that $f(AB)=f(BA)$ for each $A$ and $B$ in $W$, then $f$ is a scalar multiple of the trace function.
Proof:
*$\mathcal{B}$ is the standard ordered basis of $F^n$.
*$A^0$ is the annihilator of $A$.
Let $W_0$ be the subspace spanned by the matrices $C$ of the form $C=AB-BA$. Then the annihilator of $W_0$ is the subspace of $f$ that satisfies the condition stated above. Since $ \text{dim }W=n^2$, showing that $\text{dim }W_0=n^2-1$ proves that $f$ is a scalar multiple of the trace function because $ \text{trace } \in W^0_0$.
We know that the null space of the trace function has dimension $n^2-1$. If we show that the null space of the trace function is precisely the subspace $W_0$, then we are proving that $\text{dim }W_0=n^2-1$.
Fix $1\leq i\leq n$ and $1\leq j\leq n$ such that $i \neq j$. We already know that there are linear transformations $T_1$ and $T_2$ such that
$$
T_1\epsilon_i=\epsilon_j, T_1\epsilon_k=0 \text{ if } k\neq i\\ T_2\epsilon_i=\epsilon_i, T_2\epsilon_k=0 \text{ if } k\neq i
$$
Then $T_1T_2\epsilon_i=\epsilon_j$ and $T_2T_1\epsilon_i=0$. Also, $T_1T_2\epsilon_j=0$ and $T_2T_1\epsilon_j=0$. Therefore, $A=[T_1T_2-T_2T_1]_{\mathcal{B} }$is a matrix such that it has one non-zero entry of $A_{ij}=1$. With this, we see that $W_0$ consists of $n^2-n$ matrix 'units', i.e., matrices with exactly one non-zero entry, such that none of these units have a non-zero diagonal entries.
Now, there are linear transformations $L_1$ and $L_2$ such that
$$
L_1\epsilon_i=\epsilon_j, L_1\epsilon_k=0 \text{ if } k\neq i,\\ L_2\epsilon_j=\epsilon_i, L_2\epsilon_k=0 \text{ if } k\neq j.
$$
Then $L_1L_2\epsilon_j=\epsilon_j$ and $L_2L_1\epsilon_j=0$. However, note that $L_1L_2\epsilon_i=0$ and $L_2L_1\epsilon_i=\epsilon_i$. This implies that all the matrices $M=[L_1L_2-L_2L_1]_\mathcal{B}$ that have two diagonal entries and all other entries $0$, one of which is $-1$ and the other is $1$, are elements of $W_0$. Let the subspace of these matrices be $W_1$.
Now, let’s consider the matrices with diagonal entries of:
$$
E_1=\left \{ 1,-1,0,0,0,.. \right \}\\ E_2=\left \{ 0,1,-1,0,0,... \right \}\\ E_3=\left \{ 0,0,1,-1,0,... \right \}\\ \vdots \\ E_n=\left \{ 0,0,0,0,0,...1,-1 \right \}
$$
Let $E$ be a $n-1\times n$ matrix with row vectors $E_i$ of the above. Then the row echelon matrix of $E$ has $n-1$ non-zero rows. This implies that $E_i$s are linearly independent. Note that $\text{dim }W_1\leq n-1$. Since there are $n-1$ linearly independent vectors, $\text{dim }W_1=n-1$. So that $\text{dim }W_0=n^2-1$.
This proves that $W_0$ is precisely the null space of the trace function. Since the dimension of $W_0=n^2-1$, the dimension of its annihilator must be $1$. Therefore, linear functionals that satisfy the condition above are scalar multiples of the trace function.
Your proof seems basically correct. You might mention at the beginning that all commutators $AB-BA$ are obviously all in the kernel of the trace function (the set of traceless matrices), by well known properties of the trace. Then the proof amounts to realising all elements of the spanning set of that kernel as commutators. You take the most obvious tracesless matrices as spanning set: the elementary matrices $E_{i,j}$ with $i\neq j$ (a total of $n^2-n$ independent matrices) and $E_{i,i}-E_{i+1,i+1}$ for $1\leq i<n$ (a remaining $n-1$ matrices, that together with the previous ones form a basis of the kernel). Your proof could have been a bit shorter, but you do not need to worry too much about that.