I think the path-integral is a complete red herring here! I'll try to convince you that Wick rotation yields completely equivalent way of writing the Lagrangian in classical field theory.
Consider a classical action
$$S[x] = \int L[x(t)] dt$$
where $x:\mathbb{R} \to \mathcal{M}$ for some target manifold $\mathcal{M}$. The Lagrangian is schematically given by
$$L[x(t)] = \left(\left.\frac{dx(s)}{ds}\right|_{s=t}\right)^2-V(x(t))$$
where $V(x(t))$ is some polynomial in $x(t)$ which (critically) involves no derivatives.
Now analytically continue $x$ to a function $\tilde{x}: \mathbb{C}\to\mathcal{M}$ by defining $\tilde{x}(c) = x(|c|)$ which is obviously analytic. Relabel $\tilde{x}$ as $x$ for simplicity. Define a new variable
$$\tau = it$$
inside the integral, and substitute. (Warning: there are mathematical subtleties about complex substitutions, which should be dealt with using Jordan's lemma). Ignoring the subtleties, the resulting integral is
$$S[x] = \int L[x(-i\tau)](-i)d\tau$$
Now let's examine what happens to the Lagrangian more carefully. Looking at the first term we have
$$\left(\left.\frac{dx(s)}{ds}\right|_{s=-i\tau}\right)^2$$
Change the differentiation variable to $u = is$ and this term becomes
$$\left(i\left.\frac{dx(u)}{du}\right|_{u=\tau}\right)^2$$
Relabelling $u\to s$ we see that the first term has the same form as originally, but with $t$ replaced by $\tau$ and an extra minus sign, viz.
$$-\left(\left.\frac{dx(s)}{ds}\right|_{s=\tau}\right)^2$$
Now on to the potential term. This is much simpler because $x(-i\tau)=x(|-i\tau|)=x(\tau)$ by definition so the potential term is just
$$V(x(\tau))$$
which is of exactly the same form as originally. Now we define a Euclidean Lagrangian
$$L_E(x(\tau)) = \left(\left.\frac{dx(s)}{ds}\right|_{s=\tau}\right)^2+V(x(\tau))$$
Putting it all together we find
$$S[x] = i\int L_E[x(\tau)]d\tau$$
Finally defining
$$S_E[x] = \int L[x(t)]dt$$
we see that it's mathematically equivalent to calculate the path integral as
$$\int \mathcal{D}x\exp(iS[x]) \textrm{ or } \int \mathcal{D}x\exp(-S_E[x])$$
TL;DR: OP's title question (v7) about instantons in the Minkowski signature is physically meaningless. It is an irrelevant mathematical detour run amok. The connection to physics/Nature is established via a Wick rotation of the full Euclidean path integral, not bits and pieces thereof. Within the Euclidean path integral, it is possible to consistently expand over Euclidean instantons, but it meaningless to Wick rotation the instanton picture to Minkowski signature.
In more details, let there be given a double-well potential
$$V(x)~=~\frac{1}{2}(x^2-a^2)^2. \tag{A}$$
The Minkowskian and Euclidean formulations are connected via a Wick rotation
$$ t^E e^{i\epsilon}~=~e^{i\frac{\pi}{2}} t^M e^{-i\epsilon}.\tag{B} $$
We have included Feynman's $i\epsilon$-prescription in order to help convergence and avoid branch cuts and singularities. See also this related Phys.SE post.
I) On one hand, the Euclidean partition function/path integral is
$$\begin{align} Z^E~=~&Z(\Delta t^E e^{i\epsilon})\cr
~=~& \langle x_f | \exp\left[-\frac{H \Delta t^E e^{i\epsilon}}{\hbar}\right] | x_i \rangle \cr
~=~&N \int [dx] \exp\left[-\frac{S^E[x]}{\hbar} \right],\end{align}\tag{C} $$
with Euclidean action
$$\begin{align} S^E[x]
~=~&\int_{t^E_i}^{t^E_f} \! dt^E \left[ \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^E}\right)^2+e^{i\epsilon}V(x)\right]\cr
~=~& \int_{t^E_i}^{t^E_f} \! dt^E \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^E}\mp e^{i\epsilon}\sqrt{2V(x)}\right)^2 \cr
&\pm \int_{x_i}^{x_f} \! dx ~\sqrt{2V(x)}.\end{align}\tag{D}$$
and real regular kink/anti-kink solution$^1$
$$\begin{align} \frac{dx}{dt^E}\mp e^{i\epsilon}\sqrt{2V(x)}~\approx~&0 \cr
~\Updownarrow~ & \cr
x(t^E)
~\approx~&\pm a\tanh(e^{i\epsilon}\Delta t^E). \end{align}\tag{E}$$
Note that a priori space $x$ and time $t^E$ are real coordinates in the path integral (C). To evaluate the Euclidean path integral (C) via the method of steepest descent, we need not complexify space nor time. We are already integrating in the direction of steepest descent!
II) On the other hand, the corresponding Minkowskian partition function/path integral is
$$\begin{align} Z^M~=~&Z(i \Delta t^M e^{-i\epsilon})\cr
~=~& \langle x_f | \exp\left[-\frac{iH \Delta t^M e^{-i\epsilon}}{\hbar} \right] | x_i \rangle\cr
~=~& N \int [dx] \exp\left[\frac{iS^M[x]}{\hbar} \right],\end{align}\tag{F} $$
with Minkowskian action
$$\begin{align} S^M[x]~=~&\int_{t^M_i}^{t^M_f} \! dt^M \left[ \frac{e^{i\epsilon}}{2} \left(\frac{dx}{dt^M}\right)^2-e^{-i\epsilon}V(x)\right]\cr
~=~& \int_{t^M_i}^{t^M_f} \! dt^M \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^M}\mp i e^{i\epsilon}\sqrt{2V(x)}\right)^2 \cr
&\pm i \int_{x_i}^{x_f} \! dx ~\sqrt{2V(x)},\end{align}\tag{G}$$
and imaginary singular kink/anti-kink solution
$$\begin{align} \frac{dx}{dt^M}\mp i e^{i\epsilon}\sqrt{2V(x)}~\approx~&0 \cr
~\Updownarrow~ & \cr
x(t^M)~\approx~&\pm i a\tan(e^{-i\epsilon}\Delta t^M)\cr
~=~&\pm a\tanh(i e^{-i\epsilon}\Delta t^M). \end{align}\tag{H}$$
It is reassuring that the $i\epsilon$ regularization ensures that the particle starts and ends at the potential minima:
$$ \lim_{\Delta t^M\to \pm^{\prime}\infty} x(t^M)~=~(\pm a) (\pm^{\prime} 1). \tag{I}$$
Unfortunately, that seems to be just about the only nice thing about the solution (H). Note that a priori space $x$ and time $t^M$ are real coordinates in the path integral (F). We cannot directly apply the method of steepest descent to evaluate the Minkowski path integral. We need to deform the integration contour and/or complexify time and space in a consistent way. This is governed by Picard-Lefschetz theory & the Lefschetz thimble. In particular, the role of the imaginary singular kink/anti-kink solution (H) looses its importance, because we cannot expand perturbatively around it in any meaningful way.
--
$^1$ The explicit (hyperbolic) tangent solution (E) is an over-simplified toy solution. It obscures the dependence of finite initial (and final) time $t^E_i$ (and $t^E_f$), moduli parameters, and multi-instantons. We refer to the literature for details.
Best Answer
A Wick rotation is a mathematical trick which requires a contour integral in a complex plane $z$. It exploits the residue theorem which states the integral of a function along a closed curve as $2 \pi i$ times the sum of the residues of the function inside the curve.
If you have an integral along the real axis from $-\infty$ to $+\infty$ you can, via a semicircle, extend the integral to a closed curve covering half of the complex plane. If the integrand vanishes fast enough as $\vert z \vert \to \infty$ the contour integral outcome is the same as the integral along the real axis.
You can now rotate the contour by 90° provided that it does not pass over any pole to guaranty that the outcome does not change. This is possible if the poles are situated all in the left side or all in the right side of the half complex plane. Now the contour integral outcome is given by the the integral along the imaginary axis.
The procedure is formalized as $t$ replaced by $i \tau$.
Note: You do not rotate the real axis, but rotate the contour integral.