I know that the isometry group of $\mathbb{R}^2$ is $\text{Isom}(\mathbb{R}^2) = O(\mathbb{R}^2) \; T(\mathbb{R}^2) = \{h \circ g : h \in T(\mathbb{R}^2), g\in O(\mathbb{R}^2) \}$, where $O(\mathbb{R}^2)$ is the orthogonal group, and $T(\mathbb{R}^2)$ is the translation group.
I want to prove that $T(\mathbb{R}^2)$ is normal in $\text{Isom}(\mathbb{R}^2)$.
So, I want to show that $$r \circ T_a \circ r^{-1} \in T(\mathbb{R}^2).$$
Since all elements in $\text{Isom}(\mathbb{R}^2)$ can be written in the form $h \circ g : h \in T(\mathbb{R}^2), g\in O(\mathbb{R}^2) $, this is equivalent to proving that $$(h \circ g) \circ T_a \circ (h \circ g)^{-1} \in T(\mathbb{R}^2) .$$
where $T_a(x) = x+a$ for all $x \in \mathbb{R}^2$.
Conjugating by a translation clearly gives us a translation, so we will conjugate by an element $g \in O(\mathbb{R}^2) $
Let $x \in \mathbb{R}$ be any point in $\mathbb{R}^2$. Then, since all elements in $ \text{Isom}(\mathbb{R}^2)$ are bijections, $$(g \circ T_a \circ g^{-1} ) (x)$$ $$ = ( g \circ T_a) (g^{-1} (x)) $$ $$= g (g^{-1} (x) +a) $$ $$= (x) +g(a) $$
I am unsure of this last step. How do we know $g(x+t) = g(x) + g(t)$? What I mean is, how do we know that is isn't a homomorphism under the multiplication law, which would mean that $g(xt) = g(x) g(t)$ but wouldn't necessarily say anything about $g(x+t)$?
That isomorphism is not just an isomorphism of abstract groups, it also tells you how $(h,g)$ acts on $\mathbb{R}^2$. The function $g$ is a linear map $\mathbb{R}^2 \to \mathbb{R}^2$, i.e. a matrix that multiplies vectors from the left, therefore $g(v+u) = g(v)+g(u)$.
Maybe the confusion comes from the fact, that that what you are really studying is $\text{Isom}(\mathbb{A}^2,d)$, isometries of the affine plane with a given metric $d$. Then you can embed $O(2)$ in the group of those isometries by choosing the fixed point and the axes.
But once you choose a chart $\mathbb{R}^2 \to \mathbb{A}^2$ which respects affine structure and metric, you can see $\text{Isom}(\mathbb{A}^2,d)$ as you described it in your question.