Let $u\in C^\infty (\Omega,\mathbb R)$ be a solution of $$\begin{array}{cccl} -\Delta u & = & 0 & \mathrm{in}\ \Omega \\ u & = & g & \mathrm{on}\ \partial\Omega\end{array}$$ Whereas $\Omega\subset\mathbb R^n$. We define $\delta_\lambda(x):=\lambda x$ and $\tau_\lambda (x):= x+c$. Find PDEs and domains $\tilde\Omega_1, \tilde\Omega_2$ with solutions $u_\lambda :=u\circ \delta_\lambda$ and $u_c:=u\circ\tau_c$ respectively.
I think by introducing $x'=x+c$ we have $$\frac{\partial u(x')}{\partial x} = \frac{\partial u}{\partial x'}\frac{\partial x'}{\partial x} = \frac{\partial u}{\partial x'},$$ so $u\circ\tau_c$ is also a solution of $-\Delta u =0$? I'd say we just have to change our domain to $\tilde\Omega_1:=\Omega+c:=\{x+c|x\in\Omega\}$.
Also we set $y'=\lambda x$ and get $$\frac{\partial u(y')}{\partial x} = \frac{\partial u}{\partial y'}\frac{\partial y'}{\partial x} = \lambda \frac{\partial u}{\partial y'}$$ so $u\circ\delta_\lambda$ is a solution of $-\frac{1}{\lambda^2}\Delta u=0$ on $\tilde\Omega_2:=\lambda\Omega:=\{\lambda x|x\in\Omega\}$.
Am I going in the right direction with this?
The domains like I said in the comments above $$ u\circ \tau_c :\widetilde{\Omega}_1 \to \mathbb{R}, $$ knowing $u: \Omega\to \mathbb{R}$, we must have $$ \tau_c : \widetilde{\Omega}_1 \to \Omega, $$ this is to say, $$ \Omega = \tau_c(\widetilde{\Omega}_1), $$ applying the inverse of the translation operator $\tau_{-c}$ on both sides: $$ \widetilde{\Omega}_1 = \tau_{-c}{\Omega} = \{x'\in \mathbb{R}^n : x' = x-c\text{ for some }x\in \Omega\}. $$ Similarly $\widetilde{\Omega}_2 = \delta_{1/\lambda}{\Omega}$. Both are the same with what you wrote in the comments. Following this kind of notation will get you through more complicated coordinate change.
Also here is a comment for the differentiation you have in your question: notice in the new domain $\widetilde{\Omega}_1$, the differentiation should be with respect to $x'\in \widetilde{\Omega}_1$, not with respect to the original $x$ (just to avoid confusion of notations), the partial derivative should be: $$ \frac{\partial (u\circ \tau_c)(x')}{\partial x'} = \frac{\partial u(x)}{\partial x}\Bigg|_{x = \tau_c(x')}\cdot \frac{\partial x}{\partial x'}\Bigg|_{x = \tau_c(x')} = \frac{\partial u(x)}{\partial x}\Bigg|_{x = \tau_c(x')}, $$ the new equation should be letting $(u\circ \tau_c)(x') = u_{c}$ $$ -\Delta_{x'} u_{c} = 0, \quad \text{ in } \widetilde{\Omega}_1,\\ u_{c}(x') = (g\circ \tau_c)(x') \quad \text{ on } \partial \widetilde{\Omega}_1. $$ Now do the same thing for the translation operator.