Are those two ways to relate Extensions to Ext equivalent?

Given an extension of $R$-modules $0\to B\to X\to A \to 0$, one usually associates $x\in\operatorname{Ext}^1(A,B)$ to this extension by taking the long exact sequence

$$\dotsb\to \operatorname{Hom}(A,X) \to \operatorname{Hom}(A,A) \xrightarrow{\partial} \operatorname{Ext}^1(A,B)\to \dotsb$$

and setting $x=\partial(\mathrm{id}_A)$. Alternatively one could apply $\operatorname{Ext}^{*}(-,B)$ to get

$$\dotsb\to \operatorname{Hom}(X,B) \to \operatorname{Hom}(B,B) \xrightarrow{\partial} \operatorname{Ext}^1(A,B)\to \dotsb$$

and take $y=\partial(\mathrm{id}_B)$. Do we get the same elements in this way? I.e. is $x=y$? Optimally, can you show this from the standard properties of $\operatorname{Ext}$? I became interested in this because it seems to be necessary to solve a more particular question about a proof I had.

We can compute this in the derived category.

Extensions give distinguished triangles, and are determined by the corresponding morphism $f : A \to B[1]$.

The two methods you describe for associating an element to the extension correspond to pre- and post-composition with $f$:

$$ \hom(A, A) \xrightarrow{g \mapsto f\circ g} \hom(A, B[1]) $$ $$ \hom(B[1], B[1]) \xrightarrow{h \mapsto h\circ f} \hom(A, B[1]) $$

and so the same element $f \in \hom(A, B[1])$ is indeed obtained by applying these maps to the respective identity morphisms.

Nice observation. Actually, I don't know, if the two agree. Also, this is no answer, but a roadmap to attack the problem.

The problem is that (in the notion of Weibel) $$x = \partial(id_A) \in R^\ast Hom(-,B)(A)$$ while $$y = \partial(id_B) \in R^\ast Hom(A,-)(B)$$ So $x, y$ don't belong to the same set and can't be compared directly.

Let $P \to A$ be a projective resolution and $B \to I$ an injective resolution. Then, as in the proof of Weibel 2.7.6 there are natural isomorphisms
$$H^\ast Hom(A,I) \xrightarrow{f} H^\ast\operatorname{Tot} Hom(P,I) \xleftarrow{g} H^\ast Hom(P,B).$$

So what one could hope for is: $g(x)=f(y)$.