Searching more on the topic, I've found this post on Math SE with a much stronger result that in particular implies the existence of $T_1$ as defined above. I will reproduce it here to keep things self-contained and prevent question deletions from rendering this answer meaningless.
It is true that
Given two edge-disjoint Hamiltonian cycles, there is a different pair of edge-disjoint Hamiltonian cycles with the same union.
This follows from a result of <PERSON>, published in Section 2 of
<PERSON>, A.G., Hamiltonian Cycles and Uniquely Edge Colourable Graphs, Ann. Discrete Math. 3, 259-268 (1978).
The rest of this answer describes how that result applies.
Given a multigraph $G = (V,E)$, possibly having loops, define a Hamiltonian pair to be a pair ${h,\bar{h}}$ of edge-disjoint Hamiltonian cycles in $G$. Note that there are different conventions for counting Hamiltonian cycles of multigraphs. For our purposes, the "fat triangle" consisting of three vertices, each pair of which is joined by two edges, has four distinct Hamiltonian pairs.
A Hamiltonian cycle can be picked in $2^3$ ways (for each pair of vertices, choose one of two edges connecting them), and uniquely defines the other possible cycle in the pair (for each pair of vertices, use the edge that was not chosen).
Since which cycle was picked first doesn't matter, the result is $2^3/2 = 4$.
Let $P_G$ denote the set of Hamiltonian pairs of $G$.
For $e, e'\in E$, let $P_G(e, e')$ be the set of Hamiltonian pairs of $G$ for which $e$ and $e'$ lie in the same cycle.
Note that if $x$ is a vertex of $G$ and $G$ is $4$-regular, then $x$ is the endpoint of exactly four edges, say $e_1$, $e_2$, $e_3$, $e_4$, and $P_G = \bigsqcup_{i=2}^4 P_G(e_1,e_i)$, where the union is disjoint. 180
It follows that $|P_G| = \sum_{i=2}^4 |P_G(e_1,e_i)|$.
THEOREM (<PERSON>): If $G$ is a $4$-regular multigraph with at least three vertices and $e$ and $e'$ are two edges of $G$, then $|P_G(e, e')|$ is even.
By our earlier observation, which is made in <PERSON>'s paper, if one can prove that $|P_G(e_1, e_i)|$ is even for all $i = 2, \dots, 4$, it will follow that $|P_G|$, the number of Hamiltonian pairs, is even.
Now, let $G$ be the union of two edge-disjoint Hamiltonian cycles on $n$ vertices, viewed as subgraphs of $K_n$ (the complete graph on $n$ vertices). Then $G$ is a $4$-regular graph.
One Hamiltonian pair is given by taking the two Hamiltonian cycles defining $G$. Since one can assume $G$ has at least three vertices, using <PERSON>'s result, one can produce another Hamiltonian pair in $G$, which is necessarily a Hamiltonian decomposition of $G$ because $G$ is $4$-regular.
NOTE: At this point we already have what we need: either cycle in the second Hamiltonian pair is a valid $T_1$.
<PERSON> goes on to prove the stronger result that if $m \ge 1$ and $G$ is a $2m$-regular multigraph with at least three vertices, then $G$ has at least $(3m-2)(3m-5)...7 \cdot 4$ Hamiltonian decompositions. In particular, $4$-regular graphs have at least $4$ distinct Hamiltonian pairs.
NOTE: In our particular case, $G$ is the union of two edge-disjoint Hamiltonian cycles. This means that for any Hamiltonian cycle $C$ in $G$ there's exactly one Hamiltonian cycle that is edge-disjoint from $C$. In other words, any two distinct Hamiltonian pairs are disjoint. 433