Won't the following argument show that the difference between successive exponents can never be bounded away from zero no matter how clever you try to be in selecting $(a,b)$?

The idea is to consider pairs of integers $(n,m)$ such that $na-mb$ is close to zero.

 1. If the ratio $a/b$ is rational then we can find integers $n,m$ such that $na=mb$. But in that case
$$(n+a)^2+(-m+b)^2=n^2+m^2+a^2+b^2+2(an-mb)=n^2+m^2+a^2+b^2$$
and also
$$
(-n+a)^2+(m+b)^2=\cdots=n^2+m^2+a^2+b^2
$$
meaning that the theta series has coefficients $>1$.
 2. On the other hand if the ratio $a/b$ is irrational then, to a given $\epsilon>0$, we can find integers $n,m$ such that
$$
|na-mb|<\epsilon.
$$
This is because the additive group generated by $a$ and $b$ is then a dense subset of $\Bbb{R}$. But, reusing the above points, we see that
$$
\begin{aligned}
||(n+a,-m+b)||^2-||(-n+a,m+b)||^2&=(n^2+m^2+a^2+b^2)+2(an-mb)\\
&-(n^2+m^2+a^2+b^2)-2(an-mb)\\
&=4(an-mb),
\end{aligned}
$$
which is $<4\epsilon$.

A geometrical motivation for finding these points came from the observation that when $\pm(n,-m)$ is nearly orthogonal to $(a,b)$, we are bound to get two points yielding nearly equal exponents of $q$.