$\mathbf{R}^*$ be a group where the operation is multiplication, the group G is 2x2 matrices with non-zero determinant. Is f isomorphic and ...?

Let $$\mathbf{R}^*$$ be group of non-zero real numbers where the operation is multiplication. The group is 2x2 matrices with non-zero determinant.
Let $$f: G \rightarrow \mathbf{R}^*$$ be a function given by $$f(A)= det(A)$$.

My proof to show f is homomorphorism is:

We must show $$f(ab)=f(a)f(b)$$. Here we can see
if $$A= \begin{pmatrix} a & b\\ c & d \end{pmatrix}$$ and $$B= \begin{pmatrix} e & f\\ g & h \end{pmatrix}$$,

Then $$f(AB) = f(C)$$ where $$C = AB = \begin{pmatrix} ae+bg & af+bh\\ ce+dg & cf+dh \end{pmatrix}$$

Note $$\det (C)$$
$$=f(C)$$ $$=(ae+bg)(cf+dh) - (af+bh)(ce+dg)$$
$$=(aecf+bgcf)+(aedh+bgdh) - (afce+bhce)-(afdg+bhdg)$$
$$=bgcf+aedh - bhce-afdg$$

Also note that $$f(A)= ad-bc$$ and $$f(B)=eh-fg$$
then $$f(A)f(B) = (ad-bc)(eh-fg)=adeh-bceh+adfg-bcfg$$

Thus $$f(AB)=f(A)f(B)$$ and as such $$f$$ is a homomorphism.

Now since we showed $$f$$ is surjective, we need to show $$f$$ is injective to prove $$f$$ is a bijective mapping and thus isomorphic.

One aspect of $$2\times2$$ determinant we can use is that it produce the area of parallelogram. If we think of the properties of rotation matrices, we know their determinants all produce $$1$$. Thus $$f$$ is not injective and bijective. Meaning $$f$$ is not isomorphic.

$$f$$ is surjection, but not injection. It can be prooved very easily, giving a specific example.
$$\forall x\in\mathbb{R}^*, f\begin{pmatrix} x & 0\\ 0 & 1 \end{pmatrix}=x. f(I)=f(-I)$$
The first example shows $$f$$ is surjective, and the second shows $$f$$ is not injective.
Also, your approach to injectivity is right, but your approach to surjectivity does not show that $$f$$ is surjection.