Is this set dense in $L^2(0,1)$? What I have here is the neighborhood $U := \{z\in\mathbb C : |z| < \varepsilon\}$ of zero in $\mathbb C$ and continuous functions $a,b : U\times [0,1]\to\mathbb C$ such that $|za(t) + b(t)|\ge\delta > 0$ for all $(z,t)\in U\times [0,1]$.

My question: Is the linear span of $\{e^{zt}(za(t)+b(t)) : z\in U\}$
  dense in $L^2(0,1)$?

I know this is true for $a=0$ for if $g\in L^2$ is such that $g\perp b(t)e^{zt}$ for all $z\in U$, then
$$
0 = \int_0^1\overline{g(t)}b(t)\sum_{k=0}^\infty\frac{z^kt^k}{k!}\,dt = \sum_{k=0}^\infty\frac{a_k}{k!}z^k,
$$
where $a_k = \int_0^1\overline{g(t)}b(t)t^k\,dt$. Since $(a_k)$ is bounded, the series on the rhs defines a holomorphic function on $U$ and hence $a_k = 0$ for all $k$, which means that $bg=0$ and hence $g=0$ since $|b|\ge\delta$. But I don't know how to handle this $z$-dependent function $za+b$.

Your set will not be dense in general. 
For instance, take $a=0$, $b=\delta$. In this case you are asking if $\{e^{zt}:\ z\in U\}$ is dense. Consider $g=1_{[0,1/2]}$. Then, writing $z=x+iy$, 
\begin{align}
\|e^{zt}-g\|_2^2&=\int_0^1|e^{zt}-1_{[0,1/2]}(t)|^2\,dt
\geq\int_{1/2}^1|e^{zt}-1_{[0,1/2]}(t)|^2\,dt\\
&=\int_{1/2}^1|e^{zt}|^2\,dt
=\int_{1/2}^1|e^{xt}|^2\,dt\\
&\geq \int_{1/2}^1|1+xt|^2\,dt\\
&\geq \int_{1/2}^1|1-\varepsilon|^2\,dt\\
&\geq\frac{|1-\varepsilon|^2}2.
\end{align}
