Let $R$ be a ring and let $T := \left\{\begin{bmatrix}a & b \\ 0 & c\end{bmatrix} \in \text{Mat}_2(R) \mid a,b,c \in R\right\}$. I have shown that $T$ is a subring of $M_2(R)$ which is noncommutative for $R ≠ \{0\}$, and $$\begin{bmatrix}a & b \\ 0 & c\end{bmatrix} \in T^* \iff \{a,c\} \subseteq R^*.$$
Now it should also be true that $T^*$ is abelian if and only if $R^* = \{1\}$. Now of course you can, assuming $T^*$ is abelian, take two arbitrary elements $\begin{bmatrix}a & b \\ 0 & c\end{bmatrix}, \begin{bmatrix}x & y \\ 0 & z\end{bmatrix} \in T^*$ and derive the necessary equalities $ax = xa, ay+bz = xb+yc$ and $cz = zc$, but I don't see how that would imply that $R$ has a trivial unit group. Another attempt would be to assume that $R^* \setminus \{1\} ≠ \emptyset$, let $u \in R^* \setminus \{1\}$ and consider some products like $$\begin{bmatrix}u & 1 \\ 0 & u\end{bmatrix}\begin{bmatrix}u & 0 \\ 0 & u\end{bmatrix}.$$ Then you'd see that $u^2 = 1$, but not necessarily that $u = 1$. It seems there should be a smarter argument.
You are very close; you just aren't taking the right products.
Suppose $u\in R^*$, and consider the invertible matrices $$\left(\begin{array}{cc}u&1\\0&1\end{array}\right)\qquad\text{and}\qquad \left(\begin{array}{cc}u^{-1}&0\\0&1\end{array}\right).$$
If $T^*$ is abelian, then the two products are equal, so $$\begin{align*} \left(\begin{array}{cc} u&1\\ 0 &1\end{array}\right)\left(\begin{array}{cc} u^{-1}&0\\0&1\end{array}\right) &= \left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right)\\ \left(\begin{array}{cc} u^{-1}&0\\ 0&1\end{array}\right)\left(\begin{array}{cc} u&1\\ 0&1\end{array}\right) &= \left(\begin{array}{cc} 1 & u^{-1}\\ 0 & 1 \end{array}\right), \end{align*}$$ hence $u^{-1}=1$.