With $Ax=b$, where $A \in \mathbb{R}^{3 \times 8}, x \in \mathbb{R}^{8 \times 1}$ I can find the least squares solution $x$ for a given $b$ such that I minimize $\left\|x\right\|^2_2$. The general least squares solution (According to my understanding) is given by $A^+b$ where $A^+$ is the Moore-Penrose pseudo inverse of $A$.
However I would like to find the least squares solution $x$ such that $x \geq 0$. And if it is possible, to find a general matrix $A'$ such that $A'b = x \geq 0$ ? Maybe the latter is impossible as the sign of $A'b$ depends on the sign of $b$ which in the problem I am considering is unconstrained.
If I understand correctly, the problem you're trying to solve is \begin{aligned} \min_{\mathbf{x}}\quad & \|A\mathbf{x}-\mathbf{b}\|^2 \\ s.t. \quad & x_i\geq 0 \end{aligned} This is known as Non negative least squares and can be solved using any convex optimization algorithm. In your case the problem is of small scale ($n=8$) so in general it's easy to solve. You can use a second order solver like quadprog or even a simple first order algorithm like Projected Gradient (with or without acceleration).
As for your last question, finding $A\,:\, A^T\mathbf{b}=\mathbf{x}\geq\mathbf{0}$, what is the reason for this requirement? I'm asking since it's different than the original question - where $A, \mathbf{b}$ are the given data (not a variables) and we look for $\mathbf{x}$ that best fits this data.