$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$ I realize this is a bit late but I'd like to elaborate on Professor Metafune's very interesting and helpful comments. I'll prove how your claim 1. follows from the following:

Lemma 1 Let $\Omega$ be a $C^{1,1}$ bounded domain in $\mathbb{R}^d$, $d \geq 2$. Suppose $f \in L^1(\Omega)$ and $ u \in W_{0}^{1,1}(\Omega)$ is a weak solution to the equation $$ -\Delta u = f .$$ Then if $ 1\leq q < \frac{d}{d-1} $there is a constant $C=C(q,d)$ such that $$ \norm u_{W_{0}^{1,q}} \leq C \norm f_{L^1}.$$

Then I'll provide a proof of this Lemma that I found in Prof. Luigi Orsina's notes [Elliptic equations with measure data]

[1]: https://www.ugr.es/~edp/Talks/Luigi-Abril_2009/Curso.pdf. I'm not sure who the proof is originally due to but I suspect it is Stampacchia.

Proof of 1.

Set $ w_{n} = v_{n} -v $, so that by Lemma 1, for all $ 1< q < 2$, we have $$ \|w_{n}\|_{W^{1,q}(\Omega)} \leq C=C(q) .$$ Now by the Sobolev embedding theorem, $$ \|w_{n}\|_{L^p(\Omega)} \leq C=C(p) .$$ Indeed, if $p >2$ this follows directly from above, and otherwise it follows directly from above and Hölder's inequality. Fix a compact subset $ K$ of $\Omega$. By the Calderón–Zygmund estimate, combined with the above, $$ \|w_{n}\|_{W^{2,p}(K)} $$ $$\leq C (\| \Delta w_{n} \|_{L^p(K)} + \|w_{n}\|_{L^p(\Omega)}) \leq C=C(K,p),$$ since $f_{n}$ are uniformly bounded in $L^p_{loc}.$ By the Rellich-Kondrachev lemma , there exists $\alpha > 0 $ such that $$\|w_{n}\|_{C^{0,\alpha}(K)} \leq C .$$ Thus by Arzela-Ascoli, $w_{n}$ must converge uniformly on $K$ after passing to a subsequence. The uniform limit must be $0$ by the weak convergence to $0$. This proves 1.

Now here is the proof of Lemma 1

We denote various function spaces without the domain $\Omega$, as $\Omega$ is understood to be the domain of all functions defined here. Roughly speaking, we truncate $f$, work with solutions to the Poisson equation with this truncated $f$, and estimate the measures of level sets using the Layer-Cake formula. Define $f_{n} = f1_{\{|f|<n\}}$, so that clearly $f_{n} \in L^{\infty} $, $ \|f_{n}\|_{L^1} \leq \|f\|_{L^1} $ and $ f_{n} \to f $ in $L^1.$ Consider $u_{n}$ to be the $H_{0}^1$ solution to $$ (1) \hspace{1 cm} -\Delta u_{n} = f_{n}. $$Define $$u_{n}^k = u_{n}1_{\{|u_{n}| <k \}} + k1_{\{|u_{n} \geq k\}}.$$ Then $u_{n}^k$ is a valid test function to test equation (1) against. Since $\nabla u_{n} = \nabla u_{n}^k$ on the support of $ \nabla u_{n}^k, $ we get $$(2) \hspace{1 cm} \int_{\Omega} |\nabla u_{n}^k|^2 \leq \int_{\Omega} f_{n}u_{n}^{k}\leq k\|f\|_{L^1}. $$ Denote by $2^* $ the Sobolev conjugate to $2$, if $d \geq 3.$ If $ d = 2, $ we simply work with some $p >2$ in what follows and the argument does not change substantially. By Sobolev's inequality and (2) , $$\bigg(\int_{\Omega} (u_{n}^k)^{2^*} \bigg)^{\frac{2}{2^*}} \leq Ck\|f\|_{L^1} .$$ Set $A_{n,k} = \{ |u_{n}| > k \} . $ From the above estimate, and using that $u_{n}^k = k$ on $A_{n,k}$, we obtain $$(3) \hspace{1 cm} |A_{n,k}| \leq \big(\frac{\|f\|_{L^1}}{k}\big)^{\frac{d}{d-2}}.$$ Now fix $\lambda > 0$. Observe that $$ \{|\nabla u_n | > \lambda \} \subset A_{n,k} \cup \{ |\nabla u_{n}| > \lambda \cap |u_{n}| < k\}.$$ However using Chebyshev and (2), we get $$(4) \hspace{1 cm} |\{ |\nabla u_{n}| > \lambda \cap |u_{n}| < k\} | \leq |\{ |\nabla u_{n}^k | > \lambda \}| \leq \frac{1}{\lambda ^2}\int_{\Omega}|\nabla u_{n}^k|^2 \leq \frac{k\|f\|_{L^1}}{\lambda ^2} .$$ In total, using (3) and (4) we get the following upper bound : $$(5) \hspace{1 cm} | \{|\nabla u_n | > \lambda \} | \leq |A_{n,k}| + \frac{k \|f\|_{L^1}}{\lambda ^2} \leq \big(\frac{\|f\|_{L^1}}{k}\big)^{\frac{d}{d-2}} + \frac{k \|f\|_{L^1}}{\lambda ^2} .$$ Now we pick $k$ to make the right hand side as small as possible, we choose $$ k = \lambda ^{\frac{d-2}{d-1}} \|f\|_{L^1}^{\frac{1}{d-1}}.$$ This gives $$(6) \hspace{1 cm} | \{|\nabla u_n | > \lambda \} | \leq C\big(\frac{\|f\|_{L^1}}{\lambda}\big)^{\frac{d}{d-1}}.$$ Fix $1\leq q < \frac{d}{d-1},$ and $t > 0.$ Then using the Layer-Cake formula and (6) we get \begin{equation*} \begin{split}\| \nabla u_{n}\|_{q}^{q} &\leq \int_{\{|\nabla u_{n}| > t\}}|\nabla u_{n}|^q + \int_{\{|\nabla u_{n}| \leq t\}} |\nabla u_{n}|^q \leq \int_{\{|\nabla u_{n}| > t\}}|\nabla u_{n}|^q + t^q|\Omega| \\ &\leq 2t^q|\Omega| + q\int_{t}^{\infty}\lambda^{q-1}|\{|\nabla u_{n} > \lambda \}|d\lambda \leq 2t^q|\Omega|^q + Cq\int_{t}^{\infty}\lambda^{q-1}\|f\|_{L^1}\lambda^{\frac{-d}{d-1}}. \end{split} \end{equation*} Evaluating this latter integral, we get $$ (7) \hspace{1 cm} \| \nabla u_{n}\|_{q}^{q} \leq 2t^q + C \frac{q\|f\|_{L^1}^{\frac{d}{d-1}}}{\frac{d}{d-1}-q}\frac{1}{t^{ \frac{d}{d-1} -q}} .$$ Choosing $ t = \|f\|_{L^1},$ we arrive at $$(8) \hspace{1 cm} \| \nabla u_{n}\|_{q} \leq C(q,d)\|f\|_{L^1}.$$ Thus, $ u_{n} $ is a bounded sequence in $ W_{0}^{1,q} ,$ so by Rellich-Kondrachev up to a subsequence, we have $ u_{n} \to u'$ in $L^q$. Moreover, $ \nabla u_{n}$ are bounded in $L^q$, so up to a subsequence $ \nabla u_{n} \rightharpoonup v'$ in the weak $L^q$ topology. It is easy to see that $v'$ must be the weak derivative of $u'$. Note that $ u'$ is in $W^{1,q}$, and $u'$ is the weak limit of the $u_{n}$ in the weak $W^{1,q}$ topology. But $W_{0}^{1,q}$ is a strongly closed and convex subset of $W^{1,q}$, so it follows that $u' \in W_{0}^{1,q} .$ But now note that for all $\varphi \in C^{\infty}_{c}(\Omega) $ $$ \int_{\Omega} f\varphi = \lim \int_{\Omega}f_{n}\varphi = \lim \int_{\Omega} \langle \nabla \varphi , \; \nabla u_{n} \rangle = \int_{\Omega} \langle \nabla \varphi , \; \nabla u' \rangle , $$ so $u'$ solves the same Dirichlet problem as $u$. By a result of Lions and Dautray, since $\partial \Omega \in C^{1,1} $, we have $u = u'.$ But then by (8) and the fact that $ \nabla u_{n} \rightharpoonup \nabla u $, $ \| \nabla u\|_{q} \leq \liminf \| \nabla u_{n}\|_{q} \leq C \|f\|_{L^1} .$ By Poincare, we have the desired estimate $$ \|u\|_{W_{0}^{1,q}} \leq C \|f\|_{L^1} $$