How can $a_n$ for the trigonometric Fourier series be $0$ for a non-odd square signal?
Indeed, if the cosine coefficients ($a_n$) of the Fourier series of a function $s(t)$ are all zero, then the function is the sum of sine terms, which are all odd, implying that the function $s(t)$ is odd.

In your case, however, not all $a_n$ are zero; in particular, $a_0=2$. This means that the wave has a "DC" or mean value of $a_0/2=1$. Since $a_n=0$ for $n&gt;0$, then the function $$u(t)=s(t)-a_0/2$$ is indeed odd, and now $a_n=0$ for all $n$.