For my own benefit, I will basically transcribe Hartshorne's proof here, with some elaborations:

We are trying to prove, that in the situation of the OP, that $RF$ and $LG$ exist, and there is a functorial isomorphism $\psi : RF\cong LG[-n]$.

1. RF exists - This follows from Lemma I.4.6 and Corollary I.5.3 under the assumptions that $\mathcal{A}$ has enough injectives, and $F$ has finite cohomological dimension.

2. LG exists - It's easy to see that $P$ satisfies (i)$^*$ and (ii)$^*$ of Lemma I.4.6. To show that it satisfies (iii)$^*$ of Lemma I.4.6, we will show that if
$$0\rightarrow X^0\rightarrow X^1\rightarrow X^2\rightarrow\cdots\rightarrow X^n$$
is exact (here $n$ is the cohomological dimension of $F$), with $X^1,X^2,\ldots,X^n\in P$, then $X^0\in P$ as well. For this, we play the same game as in de Bruyn's answer. Let $K^i\subset X^i$ be the kernel of $X^i\rightarrow X^{i+1}$. Then for every $i\ge 1$ we have an exact sequence
$$0\rightarrow K^i\rightarrow X^i\rightarrow K^{i+1}\rightarrow 0$$
(where $X^i\in P$) whence, a long exact sequence
$$0\rightarrow R^0F(K^i)\rightarrow \underbrace{R^0F(X^i)}_{0}\rightarrow R^0F(K^{i+1})\rightarrow R^1F(K^i)\rightarrow \underbrace{R^1F(X^i)}_{0}\rightarrow\cdots$$
where for $i\ge 1$, $R^jF(X^i) = 0$ for all $j < n$, which implies that $R^jF(K^i)\cong R^{j-1}F(k^{i+1})$ for all $j < n,i\ge 1$. Thus, we have:
$R^0F(K^1) = 0$, $R^1F(K^1) = R^0F(K^2) = 0$, $R^2F(K^1)=R^1F(K^2)=R^0F(K^3) = 0$, and so on, showing that $R^jF(K^1) = 0$ for all $j < n$. Of course since $F$ has cohomological dimension $n$, $R^jF(K^1) = 0$ for all $j\ne n$. Since $K^1 = X^0$, this shows that $P$ satisfies (iii)$^*$.

It is also clear from the definition that $P$ satisfies (iv)$^*$ of Cor I.5.3 (w.r.t. $G = R^nF$). Thus, if $L\subset K(\mathcal{A})$ is the triangulated subcategory of complexes in $P$, then $P$ satisfies (the dual of) the conditions of Theorem I.5.1, so $LG$ exists.

3. To define the morphism $\psi : RF\cong LG[-n]$, we note that $L_{Qis}\rightarrow D(\mathcal{A})$ is an equivalence of categories, so it suffices to define $\psi$ on $L_{Qis}$, or, in fact, by Proposition I.3.4, it suffices to define $\psi$ on $L$. Thus, for every complex $X^\bullet$ of objects of $P$, we must give a morphism in $D(\mathcal{B})$
$$\psi(X^\bullet) : RF(X^\bullet)\rightarrow LG(X^\bullet)[-n]$$
such that for any morphism of complexes, there is a commutative diagram
$$LG(f)\circ\psi(X^\bullet) = \psi(Y^\bullet)\circ RF(f)$$

...to be continued later.