Is $\frac{1}{\alpha} \in \mathbb{Q}[\alpha]$ for irrational $\alpha$?

Not, it's not true for all irrational numbers $0 \lt \alpha \lt 1$. To see this, assume it's true so you get

$$\frac{1}{\alpha} = \sum_{i=0}^{n}a_i \alpha^n \tag{1}\label{eq1A}$$

for some set of $a_i \in \mathbb{Q}$. Multiply by $\alpha$ on both sides and then subtract $1$ from both sides to get

$$\sum_{i=0}^{n}a_i \alpha^{n+1} - 1 = 0 \tag{2}\label{eq2A}$$

This means $\alpha$ is a root of the polynomial

$$p(x) = \sum_{i=0}^{n}a_i x^{n+1} - 1 \tag{3}\label{eq3A}$$

However, since all of the coefficients of the terms in $p(x)$ are rational, this can only happen with $\alpha$ being an algebraic number, so it's not true for all irrational, i.e., it doesn't hold for cases where $\alpha$ is a transcendental number.

John's answer is nice, but I thought I might write an answer which emphasizes that this has nothing to do with $\mathbb{Q}$ (or irrational numbers, for that matter).

Proposition: Let $F$ be a field, and let $R$ be an $F$-algebra which is an integral domain. Fix an element $\alpha \in R^{\times}$. Then $\alpha \in (F[\alpha])^{\times}$ if and only if $\alpha$ is algebraic (integral) over $F$.

Proof: This is mostly a matter of analyzing what $F[\alpha]$ means: it is the image of the unique (universal) $F$-algebra homomorphism $\varphi \colon F[X] \to R$ sending $X$ to $\alpha$. Note, then, that $\alpha$ is algebraic over $F$ if and only if $\ker(\varphi) \neq \langle 0 \rangle$. Moreover, since $R$ is a domain, the image of $F$ is an integral domain. Hence, we see the following:

If $\ker(\varphi) = \langle 0 \rangle$, then $\alpha$ is transcendental over $F$, and the $F$-subalgebra $F[\alpha] \subset R$ is isomorphic to $F[X]$. In particular, $\alpha \notin (F[\alpha])^{\times}$, since $X$ is not a unit of $F[X]$. (It generates a maximal ideal of $F[X]$!)

If $\ker(\varphi) \neq \langle 0 \rangle$, then $\ker(\varphi)$ is a nonzero prime ideal of $F[X]$, since the image of $\varphi$ is a domain. Since $F[X]$ is a principal ideal domain, it follows that $\ker(\varphi)$ must be maximal, and so by the first isomorphism theorem, $F[\alpha]$ is a field. In particular, $\alpha \in F[\alpha] \setminus \{0\} = (F[\alpha])^{\times}$.