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.
- 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 $>1$.
- 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 $q$.