Yes, the nonzero multiples of the identity $1$ are the only invertible elements in this algebra. I am sure that someone with more expertise in Hopf algebras than I, can provide a 'high level' proof of the result. The following is a 'low level' direct argument.

For the proof, I am using some notations and basic results from Woronowicz's original paper [W].

For $\nu \in (0,1) \cup (-1,0)$, we are considering the unital $*$-algebra $A$ generated by the elements $a,b$ subject to the relations
$$a^*a + b^* b = 1 \;\; , \;\; aa^* + \nu^2 b^* b = 1 \;\; , \;\; bb^* = b^*b \;\; , \;\; ab = \nu ba \;\; , \;\; ab^* = \nu b^* a \; .$$
In [W] it is proven that the elements $a^k b^n (b^*)^m$ and $(a^*)^k b^n (b^*)^m$ form a vector space basis of $A$. Define for $k \geq 0$, the subspace $A(k) \subset A$ as the linear span of $a^k b^n (b^*)^m$ with $n,m \geq 0$. For $k \leq 0$, define $A(k)$ as the linear span of $(a^*)^k b^n (b^*)^m$ with $n,m \geq 0$. One has $A(k_1) A(k_2) \subset A(k_1 + k_2)$ for all $k_1,k_2 \in \mathbb{Z}$.

Another ingredient is the $*$-representation $\pi$ of $A$ on the Hilbert space $H = \ell^2(\mathbb{N} \times \mathbb{Z})$, also defined in [W], and given by
$$\pi(a) e_{i,j} = \sqrt{1-\nu^2} e_{i-1,j} \;\; , \;\; \pi(b) e_{i,j} = \nu^n e_{i,j+1} .$$
Write $B = A(0)$, so that $B$ is the unital and abelian $*$-subalgebra of $A$ generated by $b$. We can view $B$ as the algebra of polynomials in two variables. We will use that $B$ has no zero divisors and that the multiples of $1$ are the only invertible elements in $B$.

For every $k \in \mathbb{N}$, denote by $H_k$ the closed linear span of $e_{i,j}$ with $i \geq k$, $j \in \mathbb{Z}$.
We use the following observations: for $k \geq 0$, the range of $\pi(a^k)$ is dense in $H$, while the range of $\pi((a^*)^k)$ is dense in $H_k$. Also, every $H_k$ is an invariant subspace for $\pi(B)$ and the resulting representation of $B$ on $H_k$ is faithful for every $k \geq 0$.

Assume that $x,y \in A$ and $xy = 1$. Write $x = \sum_n x_n$ and $y = \sum_n y_n$ with $x_n,y_n \in A(n)$. Let $n_0$, $m_0$ be the largest integers with $x_{n_0} \neq 0$ and $y_{m_0} \neq 0$. Since $xy = 1$, we must have that $n_0 + m_0 \geq 0$. The component of $xy$ in $A(n_0 + m_0)$ is given by $x_{n_0} y_{m_0}$.

We claim that $n_0 + m_0 = 0$. Assume that $n_0 + m_0 > 0$. Then, $x_{n_0} y_{m_0} = 0$. Using the notation $a_k = a^k$ for $k \geq 0$ and $a_k = (a^*)^{-k}$ for $k \leq 0$, we can uniquely write $x_{n_0} = a_{n_0} P$ and $y_{m_0} = Q a_{m_0}$ with $P,Q \in B$. We get that
$$\pi(a_{n_0}) \pi(PQ) \pi(a_{m_0}) = 0 .$$
It follows that $\pi(PQ)$ is zero on $H_k$ for $k$ large enough. Hence, $PQ = 0$. This forces $P = Q = 0$, contradicting our choice of $n_0$ and $m_0$. So the claim is proven.

Similarly define the smallest integers $n_1$, $m_1$ such that $x_{n_1} \neq 0$ and $y_{m_1} \neq 0$. The same reasoning leads to $n_1 + m_1 = 0$. It follows that $n_1 = n_0$ and $m_1 = m_0$. Exchanging the roles of $x$ and $y$ if necessary, we may assume that $m_0 \geq 0$ and $n_0 = -m_0$.
This means that $x = (a^*)^{m_0} P$ and $y = Q a^{m_0}$ for some $P,Q \in B$. It follows that
$$\pi(a^*)^{m_0} \pi(PQ) \pi(a^{m_0}) = 1 .$$
If $m_0 \geq 1$, the left hand side has a nontrivial kernel. Thus, $m_0 = 0$. It then follows that $PQ = 1$, which implies that $P$ and $Q$ are a multiple of $1$. We have proven that $x$ and $y$ are multiples of $1$.