For $1<p<\infty$ and $\lambda>0$ I want to show that $\lambda-\Delta:W^{2,p}(\mathbb{R}^{n})\to L^{p}(\mathbb{R}^{n})$ is bijective.
Injectivity is obvious since if we have $\lambda-\Delta f=\lambda-\Delta g \iff \sum_{i}D^{2}_{i}f=\sum_{i}D^{2}_{i}g$ then it implies that $f=g$ almost everywhere.
Surjectivity is a little less obvious. In the case that $p=2$ it would more or less follow from the Riesz representation theorem but that is not what we have here. Suffice to say that I am stuck at this point.
You can use the Fourier transform for a nice class of functions to solve for the resolvent. \begin{align} (\lambda I-\Delta)^{-1}f & = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\frac{\hat{f}(\xi)}{\lambda+|\xi|^2}e^{ix\cdot\xi}d\xi \\ & = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\int_{0}^{\infty}e^{-t(\lambda+|\xi|^2)}dt\hat{f}(\xi)e^{ix\cdot\xi}d\xi. \end{align} This leads to a convolution problem involving the heat kernel $$ K(t,x) = \frac{1}{(4\pi t)^{n/2}}e^{-t|x|^2}. $$ You end with $$ (\lambda I-\Delta)^{-1}f= \int_{0}^{\infty}e^{-t\lambda}\left(\frac{1}{(4\pi t)^{n/2}}\int_{\mathbb{R}^{n}}e^{-|x-y|^2/4t}f(y)dy\right) dt. $$ The inner integral is heat solution operator with $L^1$ convolution kernel of unit mass for all $t > 0$: \begin{align} H(t)f & = \frac{1}{(4\pi t)^{n/2}}\int_{\mathbb{R}^{n}}e^{-|x-y|^2/4t}f(y)dy \\ & = \int_{\mathbb{R}^{n}}K(t,x-y)f(y)dy \end{align} $$ \int_{\mathbb{R}^{n}}K(t,x)dx = 1,\;\;\; t > 0. $$ By writing $K=K^{\frac{1}{p}}K^{\frac{1}{q}}$ and the above, you can show that $$ \|H(t)f\|_{L^p} \le \|f\|_{L^p} ,\;\;\; t > 0, \; f\in L^p, \; 1 \le p < \infty. $$ Therefore, if $f \in L^p$ and $\Re\lambda > 0$, the following is in $L^p$: $$ (\lambda I -\Delta)^{-1}f = \int_{0}^{\infty}e^{-t\lambda}H(t)f dt. $$ And, \begin{align} \left\|\int_{0}^{\infty}e^{-t\lambda}H(t)fdt\right\|_{p} & \le \int_{0}^{\infty}e^{-t\Re\lambda}\|H(t)f\|_{p}dt \\ & \le \int_{0}^{\infty}e^{-t\Re\lambda}dt\|f\|_p =\frac{1}{\Re\lambda}\|f\|_p. \end{align} Because this is coming out of the Fourier transform, for smooth, well-behaved functions you know that $$ (\lambda I-\Delta)\int_{0}^{\infty}e^{-t\lambda}H(t)fdt = f. $$ So then you can apply a continuity argument to prove that the closure of $\Delta$ exists in $L^p$ and has $\sigma(\Delta)$ contained in the left half plane.
Note: This is machinery used to build $C^0$ semigroup theory. The Laplace transform of the $C^0$ semigroup $H(t)=e^{t\Delta}$ is the resolvent operator, but you don't have to know that to use the same arguments. You just need the end result, which you have through the Fourier transform and a continuity argument.