Since the coefficients
$$a_k = \frac1{2\pi i}\oint_C\frac{f(z)}{(z-c)^{k+1}}\,dz$$
for the Laurent series
$$f(z)\Big|_{r\le|z|\le R} = \sum_{k=-\infty}^{\infty}a_k\cdot(z-c)^k $$
of a function $f\in\mathcal H(B(r,R))$ (i.e. a function that is holomorphic on the Annulus of radii $r\le R$), evaluated on a circle of radius $\rho\in[r,R]$,
$$\tilde a_k := \rho^k a_k = \frac1{2\pi}\int\limits_{\phi=0}^{2\pi} f(c+\rho e^{i\phi}) e^{-ik\phi}\,d\phi \\\Rightarrow f(c+\rho e^{i\phi}) = \sum_{k=-\infty}^\infty \tilde a_k\,e^{ik\phi},$$
are (up to the factor $\rho^k$) equivalent to the Fourier coefficients of a function
$$\tilde f(x):=f\Big(c+\rho e^{i\tfrac x{2\pi\rho}}\Big),\quad x\in[-\pi\rho,+\pi\rho],$$
I was wondering if there is a meaningful limit $\rho\to\infty$ for the Laurent series. For the Fourier series, the limit becomes the Fourier transform with continous "coefficients", $\tilde a_k\to \tilde a(k)\equiv \mathcal F\{\tilde f(x)\}(k)$, such that
$$\tilde f(x) = \int_{-\infty}^\infty \tilde a(k) e^{ikx}\, dk, \\\tilde a(k) = \frac1{2\pi}\int_{-\infty}^\infty \tilde f(x) e^{-ikx}\, dx.$$
So in a similar way, this limit would turn the Laurent series into a "Laurent transform",
$$f(z)\Big|_{r\le|z|} = \int_{-\infty}^\infty a(k)\cdot (z-c)^k, \\ a(k) = \lim_{R\to\infty} \frac1{2\pi}\oint\limits_{|z|=R}\frac{f(z)}{(z-c)^{k+1}}\,dz$$
(up to some factors that I probably forgot, and assuming that $f$ remains holomorphic for $R\to\infty$), describing $f(z)$ by its values at $\mathbb C$-infinity.
So, does this make sense, has it been investigated upon before and/or any applications?
In the same way that there is a discrete $\rightarrow$ continuous transition as one goes from Fourier series to transforms, one can make a continuous analogue to Taylor series in the form of the Laplace transform. This is beautifully explained by Arthur Mattuck in this video and this one, which I heartily recommend, but the gist is this:
Start with a Taylor series of the form $\sum_{n=0}^\infty a_nx^n=:A(x)$, and change the discrete variable $n=0,1,2,\ldots$ for a continuous variable $t\in(0,\infty)$, to get $\int_0^\infty a(t) x^t dt$. From this it is only a cosmetic change to define the variable $s=-\ln(x)$, which is positive and real in the interesting range for $x$, $0<x<1$, and this makes the integral look like $$\int_0^\infty a(t)e^{-st}dt=:A(s),$$ which is just a standard Laplace transform. Typically $A(s)$ will be defined in a half-plane $\text{Re}(s)>s_0$, as long as $a(t)$ is of exponential class for $t\rightarrow\infty$.
If you now want a continuous analogue of Laurent series, then all you need to do is extend your integration to minus infinity as well: $$\int_{-\infty}^\infty a(t)e^{-st}dt=:A(s).$$ This will make the typical domain for $s$ a strip of the form $s_0<\text{Re}(s)<s_1$. This corresponds to $x=e^{-s}$ being in an annulus around zero, which is also the typical domain of convergence of Laurent series.
You are also much more constrained in your choice of $a(t)$, in the same way that coefficient sequences are more restricted for Laurent series. This is in a way more serious here, because the class of 'interesting' functions which are now available is indeed noticeably smaller than the equivalent restriction in Laurent series. (In particular, things like 'folding' a sequence so that e.g. $a_n=1/(|n|!)$ become more problematic, and can cause $a(t)$ to not be analytic anymore.) This means that the transform is slightly less useful, and indeed it is much less common in the literature than the standard Laplace transform.