Determine if a language is Decidable or semi decidable
The language $L$ is not in $\text{RE}$. Consider the complement of the halting problem $\overline{Halt_{TM}} = \{ \langle M, w\rangle: \text{ $M$ does not halt on $w$}\}$ which is not in $\text{RE}$. We show that $\overline{Halt_{TM}}$ reduces to $L$ (and therefore, $L \notin \text{RE}$). From this point, I am assuming w.l.o.g that every machine has an input alphabet that contains at least three letters $a, b, $ and $c$.
A mapping reduction from $\overline{Halt_{TM}}$ to $L$ operates as follows. On input $\langle M, w\rangle$ of the reduction, the reduction outputs $\langle K\rangle$, where $K$ is a machine that operates as follows. On input $x$ of $K$, if $x = a$ or $x = b$, $K$ accepts. Otherwise, if $x\notin \{ a, b\}$, then $K$ simulates the run of $M$ on $w$. If $M$ halts on $w$, then $K$ accepts $x$. The reduction is correct. Indeed, if $M$ does not halt on $w$, then $K$ accepts the inputs $x = a$, and $x = b$, and does not halt on any other input. In particular, $K$ accepts at most two single-letter words. Conversely, if $M$ Halts on $w$, then $K$ accepts all inputs $x$, in particular, $K$ accepts at least three single letter words $a, b,$ and $c$.
Since $\text{R}\subseteq \text{RE}$, the above also proves that $L$ is not decidable. Yet, if you want to prove that $L$ is not decidable using Rice's theorem, then you note that the property $\text{P} = \{ M: \text{ $M$ accepts at most two single-letter words} \}$ is a non-trivial semantic property:

$\text{P}$ is nontrivial: it is easy to show that there are machines that accept at most two single-letter words, and there are machines that accept at least three single-letter words. I leave the details to you.

$\text{P}$ is semantic: let $M_1$ and $M_2$ be two machines with $L(M_1) = L(M_2)$. We need to show that both $M_1$ and $M_2$ are in $\text{P}$, or both are not in $\text{P}$. This is also easy as $M_1$ and $M_1$ recognise the same language, in particular, they accept the same single-letter words.