Gluing a solid torus to a solid torus with annulus inside.

I was thinking the fact that if two genus $1$ handlebodies (solid tori) are glued via an orientation preserving homeomorphism of boundaries, the resulting manifold depends only on (up to isotopy) the image of a neighborhood of a meridian.

Let $\tau$ is a Dehn twist along a meridian. According to the above fact, the resulting manifold via $\tau$ is same as the manifold obtained via the identity homeomorphism of the boundaries since they act identity on a neighborhood of a meridian.

I'm convinced by this argument but my brain (intuition/ geometric intuition?) doesn't accept the fact. Even we do a twist, they don't change the result? So I did several experiments. One of them is as follows.

Suppose we put an annulus in a handlebody of genus $1$ along a longitude. (Homology of the annulus is not zero.) Then by the identity homeomorphism, we have $S^1 \times S^2$ and the annulus is in it as it is.

If we glue handledodies by $\tau$, we should have $S^1 \times S^2$ by the fact. But it seems for me, the annulus is "twisted" along a meridian.

Is there an isotopy between these two such that a twisted annulus is mapped to non-twisted annulus?

Or If we consider a subset in handlebody, the above fact doesn't hold true anymore?

Thank you in advance.

You appear to be missing a point-set theoretic tool to do with the quotient topology.

Let's build a space by taking two spaces $X$ and $Y$, take their disjoint union, then identify a subset of $X$ with a subset of $Y$ via a homeomorphism. Let's say $A \subset X$ and $B \subset Y$ and $\phi : A \to B$ is a homeomorphism.

$$X \sqcup_\phi Y := (X \sqcup Y) / \sim$$

where the equivalence relation $\sim$ is generated by $a \in A$ is equivalent to $\phi(a) \in B$.

Easy-to-prove-fact: (1) If $\psi : Y \to Y$ is a homeomorphism such that $\psi(B) = B$, then there is a homeomorphism $X \sqcup_\phi Y \to X \sqcup_{\psi \circ \phi} Y$. The map is defined to be the identity on $X$, and $\psi$ on $Y$.

(2) If $\eta : X \to X$ is a homeomorphism such that $\eta(A) = A$, then there is a homeomorphism $X\sqcup_\phi Y \to X \sqcup_{\phi \circ \eta} Y$. The map is the identity on $Y$ and $\eta^{-1}$ on $X$.

In your case, $X=Y = S^1 \times D^2$ and $A=B=S^1 \times S^1$. $\pi_0 Homeo(S^1 \times S^1) \simeq GL_2(\mathbb Z)$ and $\pi_0 Homeo(S^1 \times D^2) \simeq \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z$, the solid torus has a mirror reflection, you can reverse the orientation of the core and you can "twist" about it.

From this perspective, the manifolds you can generate by gluing two solid tori together are controlled by the double cosets of $\pi_0 Homeo(S^1 \times D^2)$ in $\pi_0 Homeo(S^1 \times S^1)$. Using only cosets of one type tells you this reduces to studying where the meridian goes.

A less dry argument would be to first ask where the meridian goes, and then simply observe that, once you know where the meridian goes, any two extensions of the meridian embedding to a diffeomorphism of $S^1 \times S^1$ must differ by a diffeomorphism of $S^1 \times D^2$. So here you're using some actual knowledge of the surface.