$\newcommand{\al}{\alpha}\newcommand{\Ga}{\Gamma}$Let $g:=\Ga(\al)\mathcal{I}_{a+}^\al f$, so that $g=0$ on $(a,b)$. Then for any $z\in(a,b)$
\begin{equation}
\begin{aligned}
0&=\int_a^z dx\,(z-x)^{-\al}g(x) \\
&=\int_a^z dx\,(z-x)^{-\al}\int_a^x dy\,f(y)(x-y)^{\al-1} \\
&=\int_a^z dy\,f(y)\int_y^z dx\,(z-x)^{-\al}(x-y)^{\al-1} \\
&=\int_a^z dy\,f(y)\int_0^1 du\,(1-u)^{-\al}u^{\al-1} \quad \Big[u:=\frac{x-y}{z-y}\Big] \\
&=\int_a^z dy\,f(y)\,\Ga(1-\al)\Ga(\al);
\end{aligned}
\end{equation}
the interchange of the order integration is possible because $f\in L^q(a,b)$ for some $q\ge1$ and hence $f\in L^1(a,b)$.

So, $\int_a^z f=0$ for all $z\in(a,b)$ and thus $f=0$ almost everywhere (a.e.) on $(a,b)$. $\quad\Box$

*Details on the last sentence in the proof above:* Let $\mu$ be the measure over $(a,b)$ defined by the formula $\mu(A):=\int_A f$ for all Lebesgue-measurable $A\subseteq(a,b)$. Then the condition that $\int_a^z f=0$ for all $z\in(a,b)$ implies that $\mu=0$ on the algebra over $(a,b)$ generated by all intervals of the form $(a,z)$ for $z\in(a,b)$. By the uniqueness of the extension of measure, $\mu=0$ on all Borel subsets of $(a,b)$. Also, clearly $\mu=0$ on the sets of Lebesgue measure $0$. So, $\mu=0$ on all Lebesgue subsets of $(a,b)$. So, for each real $t>0$ and $A_t:=f^{-1}((t,\infty))$ we have $0=\mu(A_t)\ge t|A_t|$, where $|\cdot|$ is the Lebesgue measure. So, $|A_t|=0$ for all real $t>0$.
So, $f\le0$ a.e. on $(a,b)$. Similarly, $f\ge0$ a.e. on $(a,b)$. Thus, $f=0$ a.e. on $(a,b)$.

The condition $q<1/\al$ was not needed or used in the proof above.