I will sketch the answers to all 4 questions. In some cases they are different from the answers for $e^x$.

**The integral formula for the remainder in Taylor's theorem** is a natural starting point for explaining all these phenomena. Although the Taylor polynomials $T_k(x)$ of $e^x-1$ are initially defined only for nonnegative integers $k$, the integral formula lets one extend the definition to non-integral values of $k$, so that one can watch the zeros of $T_k(x)$ deform continuously as $k$ gradually increases from one integer to the next!

Taylor's theorem says
$$f(x) = T_k(x) + \int_0^x \frac{(x-t)^k}{k!} f^{(k+1)}(t)\, dt,$$
which for $f(x)=e^x-1$ leads to
$$T_k(x) = e^x - 1 - \int_0^x \frac{(x-t)^k}{k!} e^t \, dt$$
for all $k \in \mathbf{Z}_{\ge 1}$.
If we interpret $k!$ as $\Gamma(k+1)$, use the straight line path from $0$ to $x$, and use the negative real axis in the $x$-plane as branch cut, then the right hand side defines an analytic function of two complex variables on the region
$$\{ (k,x) \in \mathbf{C}^2 : \text{Re}(k)>-1, x \notin \mathbf{R}_{\le 0} \}.$$
This leads to implicit equations for the "stripes" in Question 2 and the "trajectory" in Question 4.

Let's now consider Question 3, about the real zeros of $T_k(x)$ for $k \in \mathbf{Z}_{\ge 1}$. Since $T_k(x)$ has nonnegative coefficients, all nonzero zeros are negative. Set $x=-y$ and $t=-u$ in $T_k(x)=0$ to obtain
$$1-e^{-y} = (-1)^k \int_0^y \frac{(y-u)^k}{k!} e^{-u} \, du.$$
If $k$ is odd, the two sides have opposite sign for $y>0$, so $T_k(x)$ has no nonzero real zeros.
If $k$ is even, then $T_k(x) \to +\infty$ as $x \to -\infty$, but $T(x)=x+\cdots<0$ for small negative $x$, so $T_k(x)$ has a negative zero, and by Rolle's theorem there cannot be more than one. Where is it, approximately? Rewrite the equation for even $k$ as
$$1-e^{-y} = \frac{y^k}{k!} \int_0^y e^{k \log(1-u/y)} e^{-u} \, du$$
Using Stirling's formula $k! \approx (k/e)^k \sqrt{2\pi k}$, we see that $y \approx k/e$, and then the integral is approximately $\int_0^\infty e^{-eu-u} \, du = 1/(e+1)$, so even more precisely,
$$y \approx (k!/(e+1))^{1/k} \approx \frac{k}{e} + \frac{1}{2e} \log k + c + o(1)$$
where $c := \frac{1}{e} \log \left( \frac{\sqrt{2\pi}}{e+1} \right)$. I have skipped the error analysis, but it is not hard. With more work, one could obtain an asymptotic expansion. The negative real zero of $T_k(x)$ is the negative of this $y$. This answers Question 3.

Finally, let's sketch an answer to Question 1, about the locations of the complex zeros. The size of $1-e^{-y}$ is about $1$ if $\text{Re}(y)$ is somewhat positive, and then the analysis is the same as above for the negative real zero, except that now we consider also the other $k^{\text{th}}$ roots with $\text{Re}(y)<0$. So these zeros are very close to a semicircle, as predicted, and the deviations are explained by the fact that the approximation $1/(e+1)$ to the integral above gets replaced by some function of $y/k$. Again, this can be made quantitative, and proved with the aid of Rouché's theorem if desired.

To understand the complex zeros with large positive real part, go back to the first integral equation for $T_k(x)$, substitute $t=x-v$, and divide by $e^x$ to get the equation
$$1 - e^{-x} = \int_0^x \frac{u^k}{k!} e^{-u} \, du.$$
To within a factor of $e^{o(k)}$, the absolute value of the integral is $\left| \frac{x^k}{k!} e^{-x} \right|$. So if we take absolute values, use Stirling's formula, and take $k^{\text{th}}$ roots, then up to a factor of $e^{o(1)}=1+o(1)$ we get
$$ 1 \approx \frac{|x|}{k/e} \left| e^{x/k} \right| $$
so $z:=x/k$ satisfies $|z e^{1-z}| \approx 1$.
Again, this can be made precise.

Above we implicitly used that the order of magnitude of $e^x-1$ is that of $e^x$ if $\operatorname{Re}(x)$ is large and positive, and of order $1$ if $\operatorname{Re}(x)$ is large and negative. In the region in between, the latter can fail when $x$ is close to an integer multiple of $2\pi i$. This leads to zeros close to the segment joining $ik/e$ and $-ik/e$, in which the integral of the previous paragraph can be small.

**Final answer to Question 1:** If we take the zeros of $T_k(x)$ and normalize by dividing by $k$, they converge to the union of the semicircle $|z|=1/e$ with $\operatorname{Re}(z)\le 0$, the segment joining $i/e$ and $-i/e$, and the part of the curve $|ze^{1-z}|=1$ with $|z| \le 1$ and $\operatorname{Re}(z) \ge 0$.

Note how this differs from Szegö's answer for the Taylor polynomials of $e^x$ mentioned in Harald Hanche-Olsen's post.