First of all, mentioning set-theoretic background is really just making this more complicated than it needs to be, so I'll ignore it. Additionally, I'll use "$\leadsto$" in place of your "$\rightarrow$" since I also want to talk about conventional limits.

It sounds like you want $t\leadsto c$ whenever $t$ "has a form" that equals or converges to $c$ (for $t,c$ appropriate terms). This can be made straightforwardly precise, but I don't think it's particularly useful. Specifically, given an $n$-ary partial function $f:\mathbb{R}^n\rightarrow\mathbb{R}$, we can consider its multivalued-and-partial extension $\hat{f}$ to $\hat{\mathbb{R}}^n$ by setting $\hat{f}(a_1,...,a_n)\leadsto b$ iff there are sequences of (non-extended!) reals $(\alpha^1_i)_{i\in\mathbb{N}}\rightarrow a_1,..., (\alpha^n_i)_{i\in\mathbb{N}}\rightarrow a_n$ such that for each $i$ the expression $\beta_i:=f(\alpha_i^1,...,\alpha_i^n)$ is defined and $(\beta_i)_{i\in\mathbb{N}}\rightarrow b$. Here, "$\rightarrow$" refers to convergence in $\hat{\mathbb{R}}$ in the usual sense.

This gives all the examples you include, as well as things like $$\sin(\pm\infty)\leadsto x\iff x\in[-1,1]$$ and $C(x)\leadsto y$ for all $x,y\in\hat{\mathbb{R}}$ where $C$ is [Conway's base $13$ function](https://en.wikipedia.org/wiki/Conway_base_13_function). Additionally, if $f$ is total on $\mathbb{R}$ then $\hat{f}$ is total on $\hat{\mathbb{R}}$ and extends $f$ (for the latter point just take $\alpha^k_i=a$), and all "value sets" $$\hat{f}[x]:=\{y: \hat{f}(x)\leadsto y\}$$ will always be **closed** for any $f$ whatsoever.

This last observation points to the right way to think about what's going on here, in my opinion: we're really just taking the topological closure of the graph of $f$ as a subset of (the appropriate power of) $\hat{\mathbb{R}}$ and declaring that to be the graph of $\hat{f}$. But I don't think we actually get anything interesting this way.