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.
@DanielRust is correct; what you have constructed is just another compact manifold whose boundary is two genus-2 surfaces, not the cylinder on that surface. Here's some more detail.
$Z$ is homotopy equivalent to the two cylinders joined by line segments rather than disks. More precisely, $(X \coprod X' \coprod I_0 \coprod I_1) / R$, where each $I_j$ is a unit interval and the relation $R$ glues $0 \in I_j$ to $(t, j) \in X$ and $1 \in I_j$ to $(t, j) \in X'$ (choosing some base point $t \in T$). Using another homotopy equivalence to collapse $X$ and $X'$ to $T$ and $I_0$ to a point, we get $T \vee T \vee S^1$.
To show that $\Bbb{Z}^2 \ast \Bbb{Z}^2 \ast \Bbb{Z}$ is different from the fundamental group of the genus-2 surface, we'll need something like MO "Free splittings of one-relator groups".
Note that your construction works similarly if you replace $T$ by any manifold $M$, and some simpler examples may be helpful. If you take $M$ to be $S^1$, the result is a twice-punctured torus; if you take $M$ to be $S^2$, the result appears to be a twice-punctured $S^2 \times S^1$.
Best Answer
This is proven for diffeomorphisms in the following paper:
Earle, C.J. and Eeels, J. "The Diffeomorphism Group of a Compact Riemann Surface". Bull. Amer. Math. Soc. 73 (1967) 557–559.
In particular, this paper proves that the space of orientation-preserving diffeomorphisms of $S^2$ that fix three points on the circle is contractible. Note that a path in the space of diffeomorphisms is precisely an isotopy.
I don't know a reference that extends this to homeomorphisms -- we would need a proof that every homeomorphism is isotopic to a diffeomorphism.