As others have pointed out, the Mellin inversion theorem is just the Fourier inversion theorem in disguise for the particular group ${\mathbb R_+}$ with invariant measure $\frac{dx}{x}$. The goal of the Fourier transform is to express a general function as a linear combination (i.e. integral) of the characters of the group, so that in this basis the operations of translations and all commuting operations will be diagonalized. For the ${\mathbb R_+}$, these characters look like $x \mapsto x^{-s}$ (the minus sign because of the normalization you chose in the question), and they are unitary (take values in the circle) for imaginary $s$ -- the operation of multiplying characters is just addition in the $s$ variable, so in the inversion formula you have the measure $ds$. There's also this funny thing about how there are $s$ with positive real part -- this is because in the "physical space" ${\mathbb R_+}$ you're always talking about distributions which are compactly supported away from $0$ when you use this transform. Let's ignore that.

Since Mellin inversion is a disguised Fourier inversion, the real question is: why is the Fourier inversion formula on ${\mathbb R}$ true? To me the most convincing answer is the following: we can decompose a general function $f(x) = \int f(y) \delta(x-y) dy$ (this is the definition of $\delta$ but you have to take approximate delta-functions to make this rigorously work like a decomposition), so if we want to express a general function as a combination of the characters $x \mapsto e^{2 \pi i \xi x}$, it suffices to consider the $\delta$ function

$\delta(x) = \int u(\xi) e^{2 \pi i \xi x} d\xi $

Now, observe that because $\delta(x)$ is invariant under multiplication by $e^{2 \pi i \eta x}$ for any $\eta$, the distribution $u(\xi)$ is translation invariant, and therefore must be constant. After you find the constant, plugging in $\delta(x) = C \int e^{2 \pi i \xi x} d\xi$ into $f(x) = \int f(y) \delta(x-y) dy$ gives the Fourier inversion formula. Complete, rigorous proofs all follow more or less these lines, but there are many flavors of how you like to phrase it. Of course, we can write the whole argument with multiplicative characters as well.

As others have pointed out, the Mellin inversion theorem is just the Fourier inversion theorem in disguise for the particular group ${\mathbb R_+}$ with invariant measure $\frac{dx}{x}$. The goal of the Fourier transform is to express a general function as a linear combination (i.e. integral) of the characters of the group, so that in this basis the operations of translations and all commuting operations will be diagonalized. For the ${\mathbb R_+}$, these characters look like $x \mapsto x^{-s}$ (the minus sign because of the normalization you chose in the question), and they are unitary (take values in the circle) for imaginary $s$ -- the operation of multiplying characters is just addition in the $s$ variable, so in the inversion formula you have the measure $ds$. There's also this funny thing about how there are $s$ with positive real part -- this is because in the "physical space" ${\mathbb R_+}$ you're always talking about distributions which are compactly supported away from $0$ when you use this transform. Let's ignore that.

Since Mellin inversion is a disguised Fourier inversion, the real question is: why is the Fourier inversion formula on ${\mathbb R}$ true? To me the most convincing answer is the following: we can decompose a general function $f(x) = \int f(y) \delta(x-y) dy$ (this is the definition of $\delta$ but you have to take approximate delta-functions to make this rigorously work like a decomposition), so if we want to express a general function as a combination of the characters $x \mapsto e^{2 \pi i \xi x}$, it suffices to consider the $\delta$ function

$\delta(x) = \int u(\xi) e^{2 \pi i \xi x} d\xi $

One interpretation of this formal idea is that the distributions $\delta(x-y)$ are just like your usual standard basis functions.

Now, observe that because $\delta(x)$ is invariant under multiplication by $e^{2 \pi i \eta x}$ for any $\eta$, the distribution $u(\xi)$ is translation invariant, and therefore must be constant. After you find the constant, plugging in $\delta(x) = C \int e^{2 \pi i \xi x} d\xi$ into $f(x) = \int f(y) \delta(x-y) dy$ gives the Fourier inversion formula. Complete, rigorous proofs all follow more or less these lines, but there are many flavors of how you like to phrase it. Of course, we can write the whole argument with multiplicative characters as well.

Edit: The above argument assumes uniqueness of the representation, but one can also remark that if there is even a single function $f(x)$ for which $\int f(x) dx \neq 0$ and which can be realized as a linear combination $\int \hat{f}(\xi) e^{2 \pi i \xi \cdot x} d\xi$, then by rescaling, renormalizing and taking a limit, we obtain $\delta(x) = C \lim_{\epsilon \to 0} \epsilon^{-1} f(x/\epsilon)$, leading formally to the formula $\delta(x) = C \int e^{2 \pi i \xi \cdot x} d\xi$. One common rigorous execution of this philosophy is performed by taking $f$ to be a Gaussian.

As others have pointed out, the Mellin inversion theorem is just the Fourier inversion theorem in disguise for the particular group ${\mathbb R_+}$ with invariant measure $\frac{dx}{x}$. The goal of the Fourier transform is to express a general function as a linear combination (i.e. integral) of the characters of the group, so that in this basis the operations of translations and all commuting operations will be diagonalized. For the ${\mathbb R_+}$, these characters look like $x \mapsto x^{-s}$ (the minus sign because of the normalization you chose in the question), and they are unitary (take values in the circle) for imaginary $s$ -- the operation of multiplying characters is just addition in the $s$ variable, so in the inversion formula you have the measure $ds$. There's also this funny thing about how there are $s$ with positive real part -- this is because in the "physical space" ${\mathbb R_+}$ you're always talking about distributions which are compactly supported away from $0$ when you use this transform. Let's ignore that.

The real question is: why is the Fourier inversion formula on ${\mathbb R}$ true? Since Mellin inversion is the same inversion in disguise. To me the most convincing answer is the following: we can decompose a general function $f(x) = \int f(y) \delta(x-y) dy$ (this is the definition of $\delta$ but you have to take approximate delta-functions to make this rigorously work like a decomposition), so if we want to express a general function as a combination of the characters $x \mapsto e^{2 \pi i \xi x}$, it suffices to consider the $\delta$ function

$\delta(x) = \int u(\xi) e^{2 \pi i \xi x} d\xi $

Now, observe that because $\delta(x)$ is invariant under multiplication by $e^{2 \pi i \eta x}$ for any $\eta$, the distribution $u(\xi)$ is translation invariant, and therefore must be constant. After you find the constant, plugging in $\delta(x) = C \int e^{2 \pi i \xi x} d\xi$ into $f(x) = \int f(y) \delta(x-y) dy$ gives the Fourier inversion formula. Complete, rigorous proofs all follow more or less these lines, but there are many flavors of how you like to phrase it. Of course, we can write the whole argument with multiplicative characters as well.

As others have pointed out, the Mellin inversion theorem is just the Fourier inversion theorem in disguise for the particular group ${\mathbb R_+}$ with invariant measure $\frac{dx}{x}$. The goal of the Fourier transform is to express a general function as a linear combination (i.e. integral) of the characters of the group, so that in this basis the operations of translations and all commuting operations will be diagonalized. For the ${\mathbb R_+}$, these characters look like $x \mapsto x^{-s}$ (the minus sign because of the normalization you chose in the question), and they are unitary (take values in the circle) for imaginary $s$ -- the operation of multiplying characters is just addition in the $s$ variable, so in the inversion formula you have the measure $ds$. There's also this funny thing about how there are $s$ with positive real part -- this is because in the "physical space" ${\mathbb R_+}$ you're always talking about distributions which are compactly supported away from $0$ when you use this transform. Let's ignore that.

Since Mellin inversion is a disguised Fourier inversion, the real question is: why is the Fourier inversion formula on ${\mathbb R}$ true? To me the most convincing answer is the following: we can decompose a general function $f(x) = \int f(y) \delta(x-y) dy$ (this is the definition of $\delta$ but you have to take approximate delta-functions to make this rigorously work like a decomposition), so if we want to express a general function as a combination of the characters $x \mapsto e^{2 \pi i \xi x}$, it suffices to consider the $\delta$ function

$\delta(x) = \int u(\xi) e^{2 \pi i \xi x} d\xi $

Now, observe that because $\delta(x)$ is invariant under multiplication by $e^{2 \pi i \eta x}$ for any $\eta$, the distribution $u(\xi)$ is translation invariant, and therefore must be constant. After you find the constant, plugging in $\delta(x) = C \int e^{2 \pi i \xi x} d\xi$ into $f(x) = \int f(y) \delta(x-y) dy$ gives the Fourier inversion formula. Complete, rigorous proofs all follow more or less these lines, but there are many flavors of how you like to phrase it. Of course, we can write the whole argument with multiplicative characters as well.