The title, basically (we do not include inverses, but I assume it would still hold if we did? But we would generate the free group and not the semigroup). We need to show that we cannot get the same matrix by two different sequences of multiplying these two matrices together. I tried finding the general form but couldn't really.
Hints preferred - please don't spoil it unless necessary. Thanks a lot in advance!
Hint. Let $a = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $b = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$. Observe that if $n$ and $m$ are positive integers, then $$ 0 < \frac{n}{n+m} < 1 < \frac{n+m}{m} $$ Now use the fact that $(n, m)a = (n, n+m)$ and $(n,m)b = (n+m,m)$ to prove that if $u, v \in \{a,b\}^*$ satisfy $(1,1)u = (1,1)v$, then $u = v$.