I fully agree that, already when writing an ODE in the form
$$
y' = \frac{y}{1+x}, \tag{1}
$$
you're already implicitly assuming that $x \neq -1$, otherwise the fraction is not defined. However, since we're dealing with an ODE (and hence implicitly assume that $y$ is continuously differentiable), the concept of a limit is intimately involved.
So, let's look at equation $(1)$ on $I := \mathbb{R} \backslash \{-1\}$. It's easy to that on $I$, the function $\eta_c(x) = c(1+x)$ is a solution of $(1)$, for any $c \in \mathbb{R}$. Now, even though $\eta_c$ is not defined at $x=-1$, the limit $\lim_{x \to -1} \eta_c(x)$ exists. That is, both left and right limits (to $x=-1$) of exist and are equal. Furthermore, by the same reasoning, the limit of the derivative ($x \to -1$) also exists.
Therefore, it's a small step to extend $\eta_c$ to $\mathbb{R}$ by including the definition $\eta_c(-1) := \lim_{x \to -1} \eta_c(x)$. This new extended function -- let's call it $y_c$ -- is continuously differentiable on the entire real line.
Since the procedure sketched above is straightforward and unambiguous, people tend to call $y_c : \mathbb{R} \to \mathbb{R}, x \mapsto c (1+x)$ 'the solution' to $(1)$, without worrying too much about the point $x=-1$.
Be that as it may, there's still something going on at $x=-1$, of course. To see what's the problem, I think it's instructive to look at the slightly rewritten ODE
$$
(1+x) y'=y. \tag{2}
$$
It's clear that on $I$, equations $(1)$ and $(2)$ are equivalent, so they give rise to the same solutions. However, if we are looking for continuously differentiable solutions to $(2)$ on the entire real line, we take the limit $x \to -1$ on both sides of equation $(2)$. As we have assumed $y$ to be continuously differentiable on $\mathbb{R}$, both the limit $\lim_{x \to -1} y(x)$ and the limit $\lim_{x \to -1} y'(x)$ exist (and are finite), so we obtain
$$
0 \times y'(-1) = y(-1),
$$
i.e. $y(-1) = 0$. Hence, we can't choose the function value at $x=-1$; every continuously differentiable solution to $(2)$ obeys $y(-1)=0$. Of course, you can still take your initial condition $y(x_0)=y_0$ anywhere else (i.e. $x_0 \in I$), and you'll get a unique, continuously differentiable solution to $(1)$.
To summarise: equation $(2)$ is singular at $x=-1$. This has the effect that every continuously differentiable solution to $(2)$ obeys $y(-1)=0$.
The solution of an initial value problem is a function which satisfies the differential equation therefore it is continuous and passes through the initial point.
The interval of existence depends on the initial condition and it extends as far as the solution curve is differentiable and continuous as the result.
In your case you have a vertical asymptote at t=1, so the left branches do not pass the vertical asymptote.
With the given initial condition, $(-\infty, 1)$ is as far as you go without losing continuity.
Best Answer
While the expression itself is defined on $\Bbb R\setminus \{0\}$, the solution of the IVP has always only a single interval as domain, here due to the initial point thus $(0,\infty)$.
For linear DE the maximal domain is always the largest interval containing the initial point where the coefficient functions and right side are continuous, and the leading coefficient is non-zero.