Yes, this is the standard Aubin-Nitsche (or duality) trick. The idea is to use the fact that $L^2$ is its own dual space to write the $L^2$-norm as an operator norm
$$\|u\|_{L^2} = \sup_{\phi\in L^2\setminus\{0\}} \frac{(u,\phi)}{\|\phi\|_{L^2}}.$$
We thus have to estimate $(u-u_h,\phi)$ for arbitrary $\phi\in L^2$. To do that, we "lift" $u-u_h$ to $H^1_0$ by considering first for arbitrary $\phi\in L^2$ the solution $w_\phi\in H^1_0$ of the dual problem
$$\label{eq1}\tag{1}
(\nabla w_\phi,\nabla v) =(\phi,v) \qquad\text{for all }v\in H^1_0.
$$
Using the standard regularity of the Poisson equation, we know that
$$ \|w_\phi\|_{H^2}\leq C \|\phi\|_{L^2}.$$
Inserting $v=u-u_h\in H^1_0$ in \eqref{eq1} and using Galerkin orthogonality for any finite element (in your case, piecewise linear) function $w_h$ yields the estimate
$$
\begin{aligned}
(\phi,u-u_h) &= (\nabla w_\phi,\nabla (u-u_h))\\
&= (\nabla w_\phi-\nabla w_h,\nabla (u-u_h))\\
&\leq C \|u-u_h\|_{H^1}\|w_\phi-w_h\|_{H^1}.
\end{aligned}
$$
Since this holds for all $w_h$, the inequality is still true if we take the infimum over all piecewise linear $w_h$.
We therefore obtain
$$\label{eq2}\tag{2}
\|u-u_h\|_{L^2} = \sup_{\phi\in L^2\setminus\{0\}} \frac{(u-u_h,\phi)}{\|\phi\|_{L^2}} \leq C \|u-u_h\|_{H^1} \sup_{\phi\in L^2\setminus\{0\}} \frac{\inf_{w_h}\|w_\phi-w_h\|_{H^1}}{\|\phi\|_{L^2}}.
$$
This is the Aubin-Nitsche-Lemma.
The next step is now to use standard error estimates for the best finite element approximation of solutions to the Poisson equation. Since $u$ is only in $H^1$, we don't get a better estimate than
$$\label{eq3}\tag{3}
\|u-u_h\|_{H^1} \leq \inf_{v_h}\|u-v_h\|_{H^1} \leq c \|u\|_{H^1} \leq C \|f\|_{H^{-1}}.
$$
But fortunately, we can use the fact that $w_\phi$ has higher regularity since the right-hand side $\phi\in L^2$ instead of $H^{-1}$. In this case, we have
$$\label{eq4}\tag{4}
\inf_{w_h}\|w_\phi-w_h\|_{H^1} \leq c h \|w_\phi\|_{H^2} \leq C h \|\phi\|_{L^2}
$$
Inserting \eqref{eq3} and \eqref{eq4} into \eqref{eq2} now yields the desired estimate.
(Note that the standard estimates require that the polynomial degree $k$ of the finite element approximation and the Sobolev exponent $m$ of the true solution satisfy $m<k+1$, so this argument doesn't work for piecewise constant ($k=0$) approximation. We also have used that $u-u_h\in H^1_0$ -- i.e., that we have a conforming approximation -- which is not true for piecewise constants.)
Since you asked for a reference: You can find a statement (even for negative Sobolev spaces $H^{-s}$ instead of $L^2$) in Theorem 5.8.3 (together with Theorem 5.4.8) in
Susanne C. Brenner and L. Ridgway Scott, MR 2373954 The mathematical theory of finite element methods, Texts in Applied Mathematics ISBN: 978-0-387-75933-3.
