This is not exactly an answer; but I've done a lot of work on "Euler's Method" for unbounded intervals. There's a bit of a preamble before I can state the theorems though. I forgo all matrix methods, and this isn't to do with numerical approximation, but rather an analytic science. We're going to keep this discussion as elementary as possible; so for that we are assuming all the functions are holomorphic; and we're ignoring some of the finer nuances of Riemann Surfaces.
Begin by defining a functor like object on holomorphic functions. Let $\mathcal{S}$ be a domain in $\mathbb{C}$; and let $\phi_j(s,z) : \mathcal{S} \times \mathbb{C} \to \mathbb{C}$ be a sequence of holomorphic functions. Then, the inner-composition of $\phi_j$ is written as,
$$
\Omega_{j=n}^m \phi_j(s,z)\bullet z = \phi_n(s,\phi_{n+1}(s,...\phi_m(s,z)))\\
$$
These objects arise fairly naturally in the study of first order difference equations (that's difference, not differential); and we can use these objects to approximate; or equal in the limit, a first order differential equation. You can limit these things to infinity, to get what is normally called an "infinite composition"; given you have nice convergence criterion. As I'm going to keep this simple, I'll just state that,
If
$$
\sum_{j=1}^\infty |\phi_j(s,z) - z| < \infty\\
$$
And this sum converges "compactly normally" on $\mathcal{S} \times \mathbb{C}$, then the function,
$$
\Phi(s,z) = \Omega_{j=1}^\infty \phi_j(s,z)\,\bullet z\\
$$
Is a holomorphic function on $\mathcal{S} \times \mathbb{C}$. This can be intuited as the limit,
$$
\Phi(s,z) = \lim_{n\to\infty} \phi_1(s,\phi_2(s,...\phi_n(s,z)))\\
$$
This is an infinite composition of the first kind, or as a discrete system. This system is great for solving first order different equations. Where if we write $\phi_j(s,z) = z + q(s-j,z)$; then the function,
$$
Q(s,z) = \Omega_{j=1}^\infty z + q(s-j,z)\,\bullet z\\
Q(s+1) - Q(s) = q(s,Q)\\
$$
With this out of the way, we can move onto infinite compositions of the second kind; or the continuous case; which is a slightly more formal construction of Euler's method. I'll keep the conversation on the real line; but these functions should definitely be entire.
Choose an interval (usually I use arcs, so we can think of this as an arc if you want) $[a,b] \subset \mathcal{S}$. For now, let's assume this interval is arbitrarily small. Take a descending partition of this interval (arc), $b=\gamma_0 > \gamma_{1} > \gamma_{2} > ... > \gamma_n = a$; with sample points $\gamma_{j} \ge \gamma_j^* \ge \gamma_{j+1}$. Let $\Delta \gamma_j = \gamma_{j} - \gamma_{j+1}$ and let $||\Delta|| = \max_{0 \le j \le n-1} |\Delta \gamma_j|$.
Now, let's assume that $\phi(s,z) : \mathcal{S} \times \mathbb{C} \to \mathbb{C}$, and is holomorphic. Then we write the "partial compositions" of the integral as,
$$
I = \Omega_{j=0}^{n-1} z + \phi(\gamma_j^*,z)\Delta\gamma_j \,\bullet z\\
$$
Where the limit as $||\Delta|| \to 0$ converges in a specific manner. This is really just Euler's method in disguise, but we've spiced it up with a Riemann-Stieltjes like construction using partitions.
Now, I'm going to skip ahead a bit; because proving this thing converges is very technical; so just bear with me when I say it converges in a nice enough manner to continue this discussion.
I like to call this thing "the compositional integral" for reasons which will appear as follows. Write,
$$
Y_{ba}(z) = \int_a^b \phi(s,z)\,ds\bullet z\\
$$
Then the following formal laws are always satisfied.
$$
\frac{d}{db} Y_{ba} = \phi(b,Y_{ba})\\
Y_{cb}(Y_{ba}) = Y_{ca}\\
Y_{aa}(z) = z\\
Y_{ba}^{-1}(z) = Y_{ab}(z)\\
$$
And additionally, we get that Leibniz substitution works. So if $u(\alpha) = a$ and $u(\beta) = b$ we get,
$$
\int_a^b \phi(s,z)\,ds\bullet z = \int_{\alpha}^\beta \phi(u(x),z)u'(x)\,dx\bullet z\\
$$
This acts as a strict generalization of Riemann-Stieltjes integration. If I take $\phi(s,z) = p(s)$ (so that it is constant in the $z$ variable) then we're reduced to the following (just plug in the above definition),
$$
\int_a^b p(s)\,ds\bullet z = z + \int_a^b p(s)\,ds\\
$$
And the aforementioned group laws just become the usual additivity properties of the integral. So with all this out of the way, we can talk about compositional integration on unbounded domains similarly to integration on unbounded domains. And this will amount to a discussion of "Euler's Method" on unbounded domains (just choose Euler's partition).
So for example; I'll write the first result which brought me to this question; and what I feel is the answer you are looking for.
Let's take,
$$
y = \int_{-\infty}^x e^{st}\,ds\bullet t\\
$$
Which is the solution to the system of equations,
$$
y' = e^{xy(x)}\\
y(-\infty) = t\\
$$
So long as you keep $t > 0$, this thing will converge. Again, the proof is very involved, but I can sketch it.
The solution to the equation,
$$
y_h(x+h) - y_h(x) = he^{xy_h(x)}\\
y(-\infty) = t\\
$$
Is always solvable for $x \in \mathbb{R}$ and $h, t>0$. And its closed form expression is given as,
$$
y_h(x,t) = \Omega_{j=1}^\infty t + he^{(x-jh)t}\,\bullet t\\
$$
This looks a lot like our integral, but it's being defined on an unbounded domain. So, to prove that our integral converges for $t>0$? We just limit $h \to 0$. As this will satisfy,
$$
\frac{y_h(x+h) - y_h(x)}{h} = e^{xy_h(x)}\\
$$
Which is the equation for the derivative if we set $h\to 0$. Now, doing this is very very difficult. And the function I just chose here is very manufactured. But, if we view "Euler's method" on unbounded domains, as compositional integration on unbounded domains; there are many more tools at your disposal. And much of them just look like integrations. Doing this in complex scenarios is very difficult though.
But, all in all;
$$
\text{`Euler's Method' on unbounded domains} = \text{compositional integrations on unbounded domains}
$$
This says nothing about speed of convergence, or how well we approximate; this is, again, an analytic solution to your problem. I hope, maybe, this helps.
Regards,