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A quick plot of the function shows that the limit is supposed to be zero:

![Log-scale plot of the function G(x) for positive x near zero, together with the upper bound used in the squeeze proof. Both curves rapidly decrease toward zero as x approaches zero from the right.](https://math.codidact.com/uploads/03esokeb4thnfyj00agizhfh8wlp)

Now we give a proof. 

Yes. The substitution $t=1/x$ is only for readability; the whole proof can be written directly in terms of $x$ and $N$.

Let

$$
N=\left\lfloor \frac1x\right\rfloor.
$$

For $0\lt x\lt 1/2$, we have $N\ge 1$, and

$$
0\le \sum_{n=N}^{\infty}\frac{x^n}{n}
\le \frac1N\sum_{n=N}^{\infty}x^n
=
\frac{x^N}{N(1-x)}.
$$

Therefore

$$
0\le e^{1/x}\sum_{n=N}^{\infty}\frac{x^n}{n}
\le
\frac{e^{1/x}x^N}{N(1-x)}.
$$

Now use only facts about $N=\lfloor 1/x\rfloor$. Since

$$
N\ge \frac1x-1,
$$

and $0\lt x\lt 1$, larger exponents make $x^a$ smaller. Hence

$$
x^N\le x^{1/x-1}.
$$

Also, for $0\lt x\lt 1/2$,

$$
N=\left\lfloor \frac1x\right\rfloor
\ge \frac1x-1
\ge \frac{1}{2x}.
$$

Thus

$$
\frac{e^{1/x}x^N}{N(1-x)}
\le
\frac{e^{1/x}x^{1/x-1}}{(1/(2x))(1-x)}.
$$

Simplifying the right-hand side gives

$$
\frac{e^{1/x}x^{1/x-1}}{(1/(2x))(1-x)}
=
\frac{2}{1-x}(ex)^{1/x}.
$$

Therefore

$$
0\le e^{1/x}\sum_{n=N}^{\infty}\frac{x^n}{n}
\le
\frac{2}{1-x}(ex)^{1/x}.
$$

Finally,

$$
(ex)^{1/x}\to 0
$$

because

$$
\log\left((ex)^{1/x}\right)
=
\frac{1+\log x}{x}\to -\infty
\qquad (x\to 0^+).
$$

Also,

$$
\frac{2}{1-x}\to 2.
$$

Hence the upper bound goes to $0$. By the squeeze theorem,

$$
\boxed{
\lim_{x\to 0^+}e^{1/x}\sum_{n=\lfloor 1/x\rfloor}^{\infty}\frac{x^n}{n}
=0.
}
$$