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This is just a calculation. Continuning your argument, $q^*(s,t) = q^*(s,0) = (s_1,s_2)$ for some $s_1\in A, s_2\in J$. Then $$ \overline{f} s - \overline{g}s = \langle (f,g), (s_1,s_2) \rangle = \overline{f} s_1 + \overline{g}s_2, $$ for all $f\in A, g\in J$. Set $f=1,g=0$ to see that $s = s_1$; set $f=0$ to see that $$ \overline{g} s = \overline{g} s_2, $$ for all $g\in J$. Letting $g$ run through an approximate identity for $J$ (so a net $(g_i)$ with $g_i(x)\rightarrow 1$ for each $x>0$) we conclude that $s(x) = s_2(x)$ for all $x>0$. If for example $s=1\in A\setminus J$ this shows that $s_2(x)=1$ for all $x>0$, contradicting that $s_2\in J$.