Let $\Omega$ be a bounded open subset of $\mathbb{R}^2$ and $w(>0)\in H_{0}^1(\Omega)$ satisfies the equation $$ -\Delta w\leq e^w\text{ in }\Omega. $$ Let $v(>0)\in H_{0}^1(\Omega)$ satisfies the equation $$ -\Delta v=e^v\text{ in }\Omega. $$ Then is it true that $w\leq v$ in $\Omega$?
I am trying to prove by observing that $w$ is a subsolution of the second equation, but unable to derive the result.
Your claim seems to be wrong.
Let $K$ be a compact subset of $\Omega$ and we set $f = \chi_K$ (characteristic function of $K$) and $w_0 := (-\Delta)^{-1}(f) \in H_0^1(\Omega)$. Then, we should have $w_0 \ge c$ on $K$ for some $c > 0$. Now, we choose $\lambda > 0$ with $\lambda < e^{\lambda \, c}$. We set $w := \lambda \, w_0$. This yields $$ -\Delta w = -\lambda \, \Delta w = \lambda \, f. $$ On $K$, we have $$ \lambda \, f = \lambda \le e^{\lambda \,c } \le e^{\lambda \, w_0} = e^{w}.$$ On $\Omega \setminus K$, we have $$ \lambda \, f = 0 \le 1 \le e^{w}.$$
Hence, $w$ solves your equation, but with $\lambda \to \infty$ (which is possible), it cannot satisfy $w \le v$ for a fixed $v$.