When does convergence in range imply convergence of arguments?

If $T$ is not injective, then choosing distinct points $a$ and $b$ in its kernel and defining a sequence $x_n$ in $H$ by the recipe $x_n = a$ if $n$ odd and $x_n = b$ if $n$ even will produce a non-convergent sequence $x_n$ in $H$ for which the sequence $Tx_n = 0$ is convergent. So it is necessary that $T$ be injective for the desired condition to hold. And if $T$ is injective, it is nonzero (outside of the trivial case $H = \{0\}$ which can be handled separately). And a nonzero linear functional is surjective. Hence $T$ is a bijection, where the argument you hint at applies.

If the codomain of $T$ is not assumed to be $\mathbb{R}$ but just a more general complete linear space (i.e., one where "$T$ nonzero" does not automatically imply "$T$ is surjective"), it appears to be necessary that $T$ be both injective and have closed range. If the range of $T$ is closed then your "bounded inverse theorem for more general codomains" (you don't state the hypotheses, but I could guess them) likely applies. If the range of $T$ is not closed, then there is a sequence $x_n$ in the unit sphere for which $T x_n \to 0$, see e.g. Closed Range (injective operator), and if the sequence $x_n$ happened to be norm convergent, you could multiply it by a sequence of unimodular scalars (e.g. $(-1)^n$) to produce examples of sequences that do not converge with the same property.