Show compactification of positive real plane is homeomorphic to $\Bbb R \Bbb P^2$

I believe there are some nice, correct ideas lurking in your first four paragraphs, but they're not very clearly expressed because of ambiguities and possible errors in your notation. However, I think you took a wrong turn when you chose to "work in a lower dimension". When you do this, what you'll end up with is a compactification not of $\ \mathbb{R}_{>0}^2\ $, but of $\ \mathbb{R}_{>0}\ $. From some of your comments, I also suspect you're harbouring some misconceptions about projective spaces, and how they can be considered compactifications of Euclidean spaces.

Note first that $\ \mathbb{R}^n_{>0}\ $ is homeomorphic to $\ \mathbb{R}^n\ $, with the generalisation $\ f_n:\mathbb{R}^n\rightarrow\mathbb{R}^n_{>0}\ $ of the $\ f\ $ you give for $\ n=2\ $, $$ f_n\big(x_1,x_2,\dots,x_n\big)=\big(e^{x_1},e^{x_2},\dots,e^{x_n}\big)\ , $$ providing a homeomorphism between them. Since $\ \mathbb{RP}^n\ $ is a compactification of $\ \mathbb{R}^n\ $, there is a homeomorphic embedding, $\ h_n:\mathbb{R}^n\rightarrow\mathbb{RP}^n\ $, of $\ \mathbb{R}^n\ $ onto a dense subset $\ h_n\big(\mathbb{R}^n\big)\ $ of $\ \mathbb{RP}^n\ $. The composite, $\ h_nf_n^{-1}\ $, is then a homeomorphic embedding of $\ \mathbb{R}^n_{>0}\ $ onto the same dense subset of $\ \mathbb{RP}^n\ $, thus showing that $\ \mathbb{RP}^n\ $ is a compactification of $\ \mathbb{R}^n_{>0}\ $.

If you take $\ \big(\mathbb{R}^{n+1}\setminus\big\{(0,0,\dots,0)\big\}\big)\big/\sim\ $ as your representation of $\ \mathbb{RP}^n\ $ (where $\ \sim\ $ is the equivalence relation defined by $\ x\sim\lambda x\ $ for any $\ \lambda\ne0\ $), then an explicit homeomorphism $\ h_n:\mathbb{R}^n\rightarrow$$\big(\mathbb{R}^{n+1}\setminus\big\{(0,0,\dots,0)\big\}\big)\big/\sim\ $ is given by \begin{align} h_n\big(x_1,x_2,&\dots,x_n\big)=\big\{x\in\mathbb{R}^{n+1}\,\big|\,x\sim\big(x_1,x_2,\dots,x_n,1\big)\,\big\}\ , \end{align} and so an explicit homeomorphism $\ h_nf_n^{-1}:\mathbb{R}^n_{>0}\rightarrow$$\big(\mathbb{R}^{n+1}\setminus\big\{(0,0,\dots,0)\big\}\big)\big/\sim\ $ is given by \begin{align} h_nf_n^{-1}\big(y_1,y_2,&\dots,y_n\big)=\big\{x\in\mathbb{R}^{n+1}\,\big|\,x\sim\big(\ln y_1,\ln y_2,\dots,\ln y_n,1\big)\,\big\}\ . \end{align} When using this representation, the complement, $\ \mathbb{RP}^n\setminus h_nf_n^{-1}\big(\mathbb{R}^n_{>0}\big)\ $, of $\ h_nf_n^{-1}\big(\mathbb{R}^n_{>0}\big)\ $ in $\ \mathbb{RP}^n\ $ is $$ \big\{\big\{\lambda x\,\big|\,\lambda\in\mathbb{R}\setminus\{0\}\big\}\,\big|\,x\in\mathbb{R}^{n+1}\setminus\big\{(0,0,\dots,0)\big\}, x_{n+1}=0\ \big\}\ , $$ which is homeomorphic to $\ \mathbb{RP}^{n-1}\ $. Thus, this embedding effectively makes $\ \mathbb{RP}^n\ $ into a disjoint union of $\ \mathbb{R}^n_{>0}\ $ and $\ \mathbb{RP}^{n-1}\ $. Since $\ \mathbb{RP}^{n-1}\ $ is not a singleton unless $\ n=1\ $, it is only for that value of $\ n\ $ that the embedding gives you a one-point compactification of $\ \mathbb{R}^n_{>0}\ $.

In fact, $\ \mathbb{RP}^n\ $ cannot be a one-point compactification of $\ \mathbb{R}^n_{>0}\ $ for $\ n\ne1\ $, because, as Ted Shifrin points out for $\ n=2\ $ in one of the comments, the one-point compactification of $\ \mathbb{R}^n\ $ $\big($or, equivalently, of $\ \mathbb{R}^n_{>0}\ \big)$ is (homeomorphic to) the $n$-sphere $\ S^n\ $, and $\ \mathbb{RP}^n\ $ is not homeomorphic to $\ S^n\ $ unless $\ n=1\ $.

As I understand it, you're trying to construct a compactification of $\ \mathbb{R}^2_{>0}\ $ by mimicing in $\ \mathbb{R}^3_{>0}\setminus\big\{(1,1,1)\big\}\ $ the construction of $\ \mathbb{RP}^2\ $ as a copy of $\ S^2\ $ with its antipodal points identified. You can certainly do this, simply by using the homeomorphism $\ f_3\ $ defined above to transfer the details of the construction from $\ \mathbb{R}^3\ $ to $\ \mathbb{R}^3_{>0}\ $, or, equivalently, from $\ \big(\mathbb{R}^3\setminus\big\{(0,0,0)\big\}\big)\big/\sim\ $ to $\ \big(\mathbb{R}^3_{>0}\setminus\big\{(1,1,1)\big\}\big)\big/f_3(\sim)\ $.

If you do this, then "the right notion of a "sphere" in $\ \mathbb{R}^3_{>0}\ $" turns out to be the surface $\ \mathbb{U}^2\ $ given by \begin{align} &\mathbb{U}^2=f_3\big(S^2\big)\\ =&\big\{\big(e^{y_1},e^{y_2},e^{y_3}\big)\,\big|\,y_1^2+y_2^2+y_3^2=1\,\big\}\\ =&\big\{\big(x_1,x_2,x_3\big)\in\mathbb{R}^3_{>0}\,\big|\,\big(\ln x_1\big)^2+\big(\ln x_2\big)^2+\big(\ln x_3\big)^2=1\,\big\}\ . \end{align} If $\ y,z\in\mathbb{R}^3_{>0}\setminus\big\{(1,1,1)\ $ then $\ f_3^{-1}\big(z_1,z_2,z_3\big)=$$\big(\ln z_1,\ln z_2,\ln z_3\big)\sim$$f_3^{-1}\big(y_1,y_2,y_3\big)=\big(\ln y_1,\ln y_2,\ln y_3\big)\ $ if and only if there is a real number $\ \lambda\ne0\ $ such that $$ \big(\ln z_1,\ln z_2,\ln z_3\big)=\lambda\big(\ln y_1,\ln y_2,\ln y_3\big)\ , $$ or, equivalently, $$ \big(z_1,z_2,z_3\big)=\big(y_1^\lambda,y_2^\lambda,y_3^\lambda\big)\ . $$ Thus, under the homeomorphism $\ f_3\ $ , the "image" $\ f_3(\sim)\ $ of the equivalence relation $\ \sim\ $ on $\ \mathbb{R}^3\setminus\big\{(0,0,0)\big\}\ $ is the equivalence relation $\ \overline{\sim}\ $ defined on $\ \mathbb{R}^3_{>0}\setminus\big\{(1,1,1)\big\}\ $ by $\ \big(y_1,y_2,y_3\big)\overline{\sim}$$\big(y_1^\lambda,y_2^\lambda,y_3^\lambda\big)\ $ for all real $\ \lambda\ne0\ $.

Two points $\ x,z\in\mathbb{U}^2\ $ will be "antipodal" to each other if $\ z\overline{\sim}x\ $—that is, $\ z_i=x_i^\lambda\ $ for some real number $\ \lambda\ne0\ $. Since $\ z\in\mathbb{U}^2\ $, then \begin{align} 1&=\big(\ln z_1\big)^2+\big(\ln z_2\big)^2+\big(\ln z_3\big)^2\\ &=\lambda^2\big(\big(\ln x_1\big)^2+\big(\ln x_2\big)^2+\big(\ln x_3\big)^2\big)\\ &=\lambda^2\ , \end{align} because $\ x\in\mathbb{U}^2\ $. Thus $\ \lambda=\pm1\ $, and $\ z\ $ will only be different from $\ x\ $ if we take $\ \lambda=-1\ $. The point on $\ \mathbb{U}^2\ $ "antipodal" to $\ x\ $ is thus the point $\ \big(x_1^{-1},x_2^{-1},x_3^{-1}\big)\ $, which I will denote by $\ x^{-1}\ $.

Thus, the compactification you're looking for is \begin{align} &\mathbb{U}^2\big/\overline{\sim}\ =\big\{\big\{x,x^{-1}\big\}\,\big|\,x\in\mathbb{U}^2\,\big\}\\ =&\big\{\big\{x,x^{-1}\big\}\,\big|\,x\in\mathbb{R}^3_{>0}, \big(\ln x_1\big)^2+\big(\ln x_2\big)^2+\big(\ln x_3\big)^2=1\,\big\}\ . \end{align}

A homeomorphism $\ \psi:\big(\mathbb{U}^2\big/\overline{\sim}\big)\rightarrow\mathbb{R}^3\setminus\big\{(0,0,0)\big\}\big)\big/\sim\ $ between this representation of $\ \mathbb{RP}^2\ $ and the more standard representation $\ \mathbb{R}^3\setminus\big\{(0,0,0)\big\}\big)\big/\sim\ $ is easy enough to obtain from a modification of $\ f_3^{-1}\ $: $$ \psi(y)=\big\{\,\lambda \big(\ln x_1,\ln x_2, \ln x_3\big)\,\big|\,\lambda\in\mathbb{R}\setminus\{0\}, x\in y\,\big\}\ . $$

What isn't quite so obvious is how to see directly that $\ \mathbb{U}^2\big/\overline{\sim}\ $ can be regarded as a compactification of $\ \mathbb{R}^2_{>0}\ $ by exhibiting an explicit homeomorphism of the latter onto a dense subset of the former. But we can do this by simply composing the mapping $\ h_2f_2^{-1}:\mathbb{R}^2_{>0}\rightarrow$$\mathbb{R}^3\setminus\big\{(0,0,0)\big\}\big)\big/\sim\ $ with the inverse of $\ \psi\ $, although the resulting expression is somewhat complicated: \begin{align} &\psi^{-1}h_2f_2^{-1}\big(y_1,y_2\big)\\ &=\left\{\left(y_1^{-d\left(y_1,y_2\right)},y_2^{-d\left(y_1,y_2\right)},e^{-d\left(y_1,y_2\right)}\right), \left(y_1^{d\left(y_1,y_2\right)},y_2^{d\left(y_1,y_2\right)},e^{d\left(y_1,y_2\right)}\right)\right\}\ , \end{align} where $\ d\big(y_1,y_2\big)=\frac{1}{\sqrt{(\ln y_1)^2+(\ln y_2)^2 +1}}\ $.

The image $\ \psi^{-1}h_2f_2^{-1}\big(\mathbb{R}^2_{>0}\big)\ $ of $\ \mathbb{R}^2_{>0}\ $ under this mapping is $$ \psi^{-1}h_2f_2^{-1}\big(\mathbb{R}^2_{>0}\big)=\big\{\big\{x,x^{-1}\big\}\,\big|\,x\in\mathbb{U}^2, x_3\ne1\,\big\}\ , $$ whose complement $$ \big\{\big\{x,x^{-1}\big\}\,\big|\,x\in\mathbb{U}^2, x_3=1\,\big\} $$ in $\ \mathbb{U}^2\big/\overline{\sim}\ $ is homeomorphic to $\ \mathbb{RP}^1\ $, and hence to a circle.