First, observe that an element $\tau$ of $\Omega(A)$ is necessarily continuous with operator norm $\|\tau\| \leq 1$. Indeed, let $x \in A$. Since $\tau(x - \tau(x)1) = 0$, the element $x - \tau(x)1$ lies in the maximal ideal $\ker\tau$, and hence cannot be invertible. In other words, $\tau(x)$ is in the spectrum of $x$, so that $|\tau(x)| \leq \|x\|$, and hence $\|\tau\| \leq 1$ — otherwise, one could use the Neumann identity to construct $(x - \tau(x) 1)^{-1}$. In fact, $\|\tau\|=1$, since $\tau(1) = 1$.
Not, let $\sigma$, $\tau \in \Omega(A)$ — a priori, then, $\sigma(1) = 1 = \tau(1)$ — and suppose that $\hat{a}(\sigma) = \hat{a}(\tau)$, which is just to say that $\sigma(a) = \tau(a)$. Because $\sigma$ and $\tau$ agree on the generators $1$ and $a$ of $A$, they agree on all polynomials in $1$ and $a$. Since such polynomials are dense in $A$, precisely by definition of $A$, it follows by continuity of $\sigma$ and $\tau$ that they agree everywhere, i.e., that $\sigma = \tau$.