The answer is **no** in general because $T$ cannot be a $GE_2$-ring in the sense of P. M. Cohn if $R$ isn't.

Let us recall some definitions. Let $n \ge 2$. A ring $\Lambda$ (not necessarily commutative) is said to be a $GE_n$-ring if the identity $GL_n(\Lambda) = D_n(\Lambda) E_n(\Lambda)$ holds. Here $D_n(\Lambda)$ denotes the group of invertible diagonal matrices over $\Lambda$ while $E_n(\Lambda)$ is the subgroup of $GL_n(\Lambda)$ generated by matrices obtained from the identity matrix by replacing an off-diagonal entry by some $\lambda \in \Lambda$.

OP's question can be rephrased in the following way (use Whitehead lemma).

**Question.** Let $R$ be commutative ring. Is the ring $T$ of $2$-by-$2$ upper triangular matrices over $R$ a $GE_2$-ring?

The answer is *no* in general.

**Claim 1**. If $T$ is a $GE_n$-ring, then so is $R$.

*Proof.* The natural surjective ring homomorphism $T \twoheadrightarrow R^2$ induces a surjective group homomorphism $\varphi_n: GL_n(T) \twoheadrightarrow GL_n(R^2)$ for every $n \ge 1$. Since $\varphi_n$ maps elementary matrices to elementary matrices, and diagonal matrices to diagonal matrices, the ring $R^2$ is certainly a $GE_n$-ring if $T$ is. It is straightforward to check that $R^2$ is a $GE_n$-ring if and only if $R$ is.

Let $k$ be field and let $m > 2$. Let $R = k[t_1, t_1^{-1}, \dots, t_m, t_m^{-1}]$. Then $R$ is not a $GE_2$-ring [4] (although $SK_1(R) = 1$ by Suslin's Stability Theorem [6, Theorem 5.8]), so that $T$ is not a $GE_2$-ring.

Below lie the remains of my first (overly complicated) answer. it has been groomed so as to address the Euclidean case $m = 1$.

**Claim 2**. Assume that $R$ is a Noetherian commutative ring of Krull dimension at most $1$ (e.g., a principal ideal domain). Then $T$ is a $GE_2$-ring if and only if $R$ is.

*Proof.*
Because of Claim $1$, we only need to prove the "if" part. To do so, let us assume that $R$ is a $GE_2$-ring. Since the Krull dimension of $R$ is at most $1$,
the Bass stable rank of $R$ is at most $2$ by Bass's Cancellation Theorem. Therefore $GL_n(R) = D_n(R) E_n(R)$ holds for all $n \ge 2$, which implies $SK_1(R) = 1$. As $K_1(T) \simeq K_1(R)^2$ by [2, Corollary 2], we deduce that $K_1(T) \simeq R^{\times} \times R^{\times}$ where the isomorphism is induced by the natural map $T^{\times} \twoheadrightarrow R^{\times} \times R^{\times}$.
Since the Krull dimension of $T$ is at most the Krull dimension of $R$, hence bounded above by $1$, the surjective stability of $K_1$ [5, Theorem 10.15] yields the result.

Note that there exists principal ideal domains which are not $GE_2$-rings, e.g., $R = \mathbf{Z}[\frac{1 + \sqrt{-19}}{2}]$ by [1, Theorem 6.1].

The following claim is straightforward.

**Claim 3.** If the stable rank $\text{sr}(R)$ of $R$ is $1$, then $\text{sr}(T) = 1$. In particular, $T$ is a $GE_2$-ring whenever $\text{sr}(R) = 1$.

I conclude with a comment on OP's motivation.
The situation of $M_2(R)$ mentioned in OP's question, is rather different than the situation of $T$. Indeed, note that $K_1(M_n(R)) = K_1(R)$ and $\text{sr}(M_n(R)) = \lceil \frac{\text{sr}(R) - 1}{n} \rceil + 1$ [5, Theorem 5.18] whereas $K_1(T_n(R)) = K_1(R)^n$ [2, Corollary 2] and $\text{sr}(T_n(R)) = 1$ if and only if $\text{sr}(R) = 1$, where $T_n(R) \subset M_n(R)$ is the ring of $n$-by-$n$ upper triangular matrices. (In some $K$-theoretical sense, $T$ is closer to $R^2$ than $M_2(R)$.)

[1] P. M. Cohn, "On the structure of the $GL_2$ of a ring", 1966.

[2] K. Dennis and S. Geller, "$K_i$ of the upper triangle matrice rings", 1976.

[3] S. Bachmuth, H. Y. Mochizuki, "$E_2 \neq SL_2$ for most Laurent polynomial rings", 1982.

[4] J. MacDonnell and J. Robson, "Noncommutative Noetherian rings", 1987.

[5] B. Magurn, "An algebraic introduction to K-theory", 2002.

[6] T. Liam, "Serre's problem on projective modules", 2006.