= Solution
Let $L_A$ and $R_A$ be left and right multiplication by $A$ on the <vector space> $M_n(\mathbb C)$. Then $\operatorname{ad}A=L_A-R_A$. On <matrix units>, or from $M_n(\mathbb C)=\mathbb C^n\otimes(\mathbb C^n)^*$, one obtains
$$
\begin{aligned}\operatorname{Tr}(L_AL_B)&=n\operatorname{tr}(AB),&\operatorname{Tr}(R_AR_B)&=n\operatorname{tr}(BA),\\\operatorname{Tr}(L_AR_B)&=\operatorname{tr}A\operatorname{tr}B,&\operatorname{Tr}(R_AL_B)&=\operatorname{tr}A\operatorname{tr}B.\end{aligned}
$$
For example, the coefficient of $E_{ij}$ in $AE_{ij}B$ is $A_{ii}B_{jj}$; summing over $i,j$ proves the mixed-trace identity. Expanding the product of the two adjoint operators therefore gives the <Killing form of the general linear Lie algebra>:
$$
\operatorname{Tr}_{M_n}(\operatorname{ad}A\operatorname{ad}B)=2n\operatorname{tr}(AB)-2\operatorname{tr}A\operatorname{tr}B.
$$
For $A,B$ in the <special linear Lie algebra>, both individual traces vanish. Furthermore $M_n(\mathbb C)=\mathfrak{sl}_n\oplus\mathbb CI$, and every adjoint operator kills the scalar summand, so its product has the same <trace> on $M_n$ and on $\mathfrak{sl}_n$. Consequently the <Killing form of the special linear Lie algebra> is
$$
\boxed{(A,B)_{\mathrm{ad}}=2n\operatorname{tr}(AB),\qquad\lambda=\frac1{2n}\quad(n\geq2).}
$$
No uniqueness theorem for <invariant bilinear forms on a Lie algebra> is needed: this calculation establishes the constant directly. For the degenerate case $n=1$, the Lie algebra is zero, both forms vanish, and proportionality does not determine a unique constant.
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