I think the answer truly does depend on whether you use the naive truncation, as in Iwaniec's book, or the Arthur truncation, as in Paul Garrett's note. In particular, the method of proof for the Maaß-Selberg relation with the naive truncation is proved in Iwaniec's book using Green's identity, and as GH from MO mentioned in his comment, the geometric picture here suggests that the proof only requires that $T \geq \sqrt{3}/2$.
On the other hand, I believe that with the Arthur truncation, one really does require that $T \geq 1$, with the proof of the Maaß-Selberg relation now instead via an unfolding argument as in Paul Garrett's notes. Here the Arthur truncation is
\[\Lambda^T E(z,s) = E(z,s) - \sum_{\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) > T}} c_0 E(\gamma z,s),
\]
where for $z = x + iy \in \mathbb{H}$,
\[c_0 E(\gamma z,s) = \int_{0}^{1} E(\gamma z,s) \, dx,\]
so that $c_0 E(z,s) = y^s + \varphi(s) y^{1-s}$.
One can easily show that if $\gamma = \begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z})$ with $\gamma \notin \Gamma_{\infty}$, so that $c \neq 0$, then
\[\Im(z) \Im(\gamma z) = \frac{y^2}{(cx + d)^2 + c^2 y^2} \leq 1,\]
while $\Im(\gamma z) = \Im(z)$ for $\gamma \in \Gamma_{\infty}$.
So for $T \geq 1$, the Arthur truncation is equal to
\[\Lambda^T E(z,s) = \begin{cases}
E(z,s) & \text{if $1/T \leq y \leq T$,} \\\
E(z,s) - y^s - \varphi(s) y^{1-s} & \text{if $y > T$.}
\end{cases}\]
If $y < 1/T$, then there may be more terms (and which terms are also present will now depend on $x$ as well), as there may be more coset representatives $\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z})$ other than just the identity for which $\Im(\gamma z) > T$.
If $T < 1$, then $\Lambda^T E(z,s) = E(z,s) - y^s - \varphi(s) y^{1-s}$ if $y \geq 1/T$, but there may be more terms if $y < 1/T$.
In particular, the Arthur truncation coincides with the naive truncation on the standard fundamental domain if $T \geq 2/\sqrt{3}$, but if $T < 2/\sqrt{3}$ then this is no longer the case.
Now a key step in proving the Maaß-Selberg relation is showing that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \left(\overline{\Lambda^T E(z,r)} - \overline{E(z,r)}\right) \, d\mu(z) = 0.\]
One shows this by unfolding to find that this is
\[-\int_{T}^{\infty} \overline{c_0 E(z,r)} \int_{0}^{1} \Lambda^T E(z,s) \, dx \, dy,\]
using the fact that $c_0 E(z,r) = y^r + \varphi(r) y^{1-r}$ does not depend on $x$.
If $T \geq 1$, then $\Lambda^T E(z,s) = E(z,s) - c_0 E(z,s)$ for $T < y < \infty$ and $0 < x < 1$, and so the inner integral vanishes. But if $T < 1$, then there may be other terms, and so this is no longer the case. So the condition $T \geq 1$ truly is necessary here.
EDIT: With regards to your comment, note that the Maaß-Selberg relation states that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \overline{\Lambda^T E(z,r)} \, d\mu(z) = \frac{T^{s + \overline{r} - 1}}{s + \overline{r} - 1} + \overline{\varphi(r)} \frac{T^{s - \overline{r}}}{s - \overline{r}} + \varphi(s) \frac{T^{\overline{r} - s}}{\overline{r} - s} + \varphi(s) \overline{\varphi(r)} \frac{T^{1 - s - \overline{r}}}{1 - s - \overline{r}}.
\]
Here
\[\varphi(s) = \frac{\Lambda(2 - 2s)}{\Lambda(2s)}, \qquad \Lambda(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s),\]
so that $|\varphi(1/2 + it)|^2 = 1$ for all $t \neq 0$, and taking logarithmic derivatives shows that
\[\frac{\varphi'}{\varphi}\left(\frac{1}{2} + it\right) = -2\Re\left(\frac{\Gamma'}{\Gamma}\left(\frac{1}{2} + it\right)\right) - 4\Re\left(\frac{\zeta'}{\zeta}(1 + 2it)\right) + 2 \log \pi.\]
Setting $s = r = 1/2 + it + \varepsilon$ for $t \neq 0$ and $\varepsilon > 0$, taking the limit as $\varepsilon \to 0$, and using the Laurent expansions of each term, we find that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \left|\Lambda^T E\left(z,\frac{1}{2} + it\right)\right|^2 \, d\mu(z) = 2 \log T - \frac{\varphi'}{\varphi}\left(\frac{1}{2} + it\right) - \Im \left( \varphi\left(\frac{1}{2} + it\right) \frac{T^{-2it}}{t}\right).
\]
Now it's not obvious to me either that this is nonnegative, and of course if $T$ is very close to zero then you wouldn't expect it to be. But for $T$ near $1$ (because as mentioned earlier, even using the naive truncation we can't have $T < \sqrt{3}/2$) it's not obvious to me either whether this is negative, so I'm not sure if there's truly a problem here; certainly for large $t$, the digamma function should dominate. Of course, I could be missing something here.
SECOND EDIT: I still don't understand your objection; as I have already stated, one requires that $T \geq 1$ for the Arthur truncation, and this is also sufficient (i.e. with the Arthur truncation, the Maaß-Selberg relation holds for all $T \geq 1$). Indeed, I already showed that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \left(\overline{\Lambda^T E(z,r)} - \overline{E(z,r)}\right) \, d\mu(z) = 0\]
when $T \geq 1$, so it remains to show that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \overline{E(z,r)} \, d\mu(z) = \frac{T^{s + \overline{r} - 1}}{s + \overline{r} - 1} + \overline{\varphi(r)} \frac{T^{s - \overline{r}}}{s - \overline{r}} + \varphi(s) \frac{T^{\overline{r} - s}}{\overline{r} - s} + \varphi(s) \overline{\varphi(r)} \frac{T^{1 - s - \overline{r}}}{1 - s - \overline{r}}.
\]
By the definition of the Arthur truncation, the left-hand side is
\[ \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \sum_ {\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) \leq T}} \Im(\gamma z)^s \overline{E(z,r)} \, d\mu(z) - \varphi(s) \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \sum_{\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) > T}} \Im(\gamma z)^{1-s} \overline{E(z,r)} \, d\mu(z),
\]
which, by the $\mathrm{SL}_2(\mathbb{Z})$-invariance of $E(z,r)$, unfolds to
\[ \int_{0}^{T} y^s \int_{0}^{1} \overline{E(z,r)} \, \frac{dx \, dy}{y^2} - \varphi(s) \int_{T}^{\infty} y^{1-s} \int_{0}^{1} \overline{E(z,r)} \, \frac{dx \, dy}{y^2}.
\]
By the definition of the constant term of $E(z,r)$ (namely that it is $y^r + \varphi(r) y^{1-r}$), one can evaluate these integrals, with the result being the Maaß-Selberg relation.
Of course, to ensure convergence of all the integrals involved, this is initially only valid for $\Re(s), \Re(r) > 1$ with $\Re(s) - \Re(r) > 1$, but then by analytic continuation this extends to all $s,r \in \mathbb{C}$ with $s \neq \overline{r}$ and $s + \overline{r} \neq 1$ provided $\varphi(s), \varphi(r)$ are well-defined. As before, one can extend this to $s = r = 1/2 + it$ with $t \neq 0$ by setting $s = r = 1/2 + it + \varepsilon$ and taking the limit as $\varepsilon \to 0$.
All this is valid if $T \geq 1$, so one certainly should be able to take $T = 1$ in the Maaß-Selberg relation, unless I'm missing something.
With regards to the integral
\[\int_{-\infty}^{\infty} h(r) \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \left|\Lambda^T E\left(z,\frac{1}{2} + ir\right)\right|^2 \, d\mu(z) \, dr,\]
to me it looks like there may be an issue at $r = 0$, and one must be careful when breaking up the integral and blindly using Weil's explicit formula. (In particular, I don't see why you say the second line is negligible - doesn't the integrand blow up at $r = 0$?) But I haven't looked too closely at this.