Is the subset of functions for some fixed $b \in \mathbb{R}^+$:
$$\mathcal{M}_{b} = \operatorname{span}\left\{ g_{\theta}(z) = \frac{\mathbb{1}_{[a,b] \cap [0,\theta]}(z) }{\sqrt{\theta -z}}, \quad \forall \theta \geq b \right\}$$
dense in $L^1[a,b]$?
Edit: I changed the question and made it more restrictive, as I figured out that the previous one had negative answer.
Let $a=0$ and $b=1.$ Denote $u=\theta^{-1},$ for $\theta>1.$ Then $$\theta^{1/2}g_\theta(x)=\left ( 1-{x\over \theta}\right )^{-1/2}=(1-ux)^{-1/2}=\sum_{n=0}^\infty a_nu^nx^n,\quad 0<u<1 $$ where $$ a_1=1,\quad a_n={-{1\over 2}\choose n}(-1)^n>0$$ and the series is convergent uniformly on $0\le x\le 1$ for any fixed $0<u<1.$ Assume for a contradiction that the linear span of $g_\theta,\theta>1,$ is not dense in $L^1(0,1).$ Then there exists a nonzero function $f\in L^\infty(0,1)$ such that $$\int\limits_0^1f(x)g_\theta(x)\,dx =0,\quad \theta >1$$ Hence $$\sum_{n=0}^\infty a_nu^n\int\limits_0^1f(x)x^n\,dx =0$$ It was possible to change the order of the infinite sum and the integral as the series is uniformly convergent. Let $m_n=\int\limits_0^1f(x)x^n\,dx.$ Then $$\sum_{n=0}^\infty a_nm_n u^n=0,\quad 0<u<1$$ As the left hand side represents an analytic function its MacLaurin coefficients must vanish, i.e. $m_n=0$ for any $n,$ as $a_n>0.$ Thus $$\int\limits_0^1f(x)p(x)\,dx=0$$ for any polynomial $p(x).$ Since the polynomials are dense in $C[0,1],$ hence in $L^1(0,1),$ we may conclude that $f=0,$ a contradiction.