If I understand correctly what you are asking then the answer is: "No".

Here's where I may be misunderstanding: I assume that $\Delta t$ is fixed. If this is correct, we can argue as follows.

Let me write $r = \Delta t$ since it is fixed and I want to disassociate it from $t$. We consider the operator $A_r \colon C^\infty_c(\mathbb{R}) \to C^\infty_c(\mathbb{R})$ defined by

$$
A_r(\phi)(t) = \int_{t - r}^{t + r} \phi(\tau) d \tau
$$

We want to extend this function to the space of distributions, $\mathcal{D} = C^\infty_c(\mathbb{R})$. To do this, we look for an adjoint as per the nlab page on distributions (particularly the section operations on distributions; note that my notation is chosen to agree with that page so it's hopefully easy to compare). So for two test functions, $\phi, \psi \in C^\infty_c(\mathbb{R})$ we calculate as follows:

$$
\begin{array}{rl}
\langle \psi, A_r(\phi)\rangle &= \int_{\mathbb{R}} \psi(t) A_r(\phi)(t) d t \\
&= \int_{\mathbb{R}} \psi(t) \int_{t - r}^{t + r} \phi(\tau) d \tau d t \\
&= \int_{\mathbb{R}} \int_{t - r}^{t + r} \psi(t) \phi(\tau) d \tau d t \\
&= \int_{\mathbb{R}} \int_{\tau - r}^{\tau + r} \psi(t) \phi(\tau) d t d \tau \\
&= \int_{\mathbb{R}} A_r(\psi)(\tau) \phi(\tau) d \tau \\
&= \langle A_r(\psi), \phi \rangle
\end{array}
$$

When we do the switch in order of integration, it's useful to draw the region of integration in the plane (I used a table "cloth" in a restaurant here in Copacabana!). It's a diagonal swathe and looks the same when flipped about the line $x = y$ (don't do this with a table cloth unless you've taken the plates off it first.).

So $A_r$ is self-adjoint and we extend it to distributions by the formula $\langle A_r(T), \phi \rangle = \langle T, A_r(\phi)\rangle$.

Now we come to your question. You have $(T_\lambda)$ such that $\langle T_\lambda, A_r(\phi)\rangle \to \langle T, A_r(\phi) \rangle$ for each test function $\phi$. Using the definition of the extension of $A_r$, this is the same as saying that $(A_r(T_\lambda)) \to A_r(T)$ weakly. You want to know if this implies that $(T_\lambda) \to T$.

To answer this, we consider $A_r$. Unfortunately for you, it has a non-trivial kernel. Note that we are now working with distributions, so can allow things with non-compact support (otherwise it wouldn't have a non-trivial kernel). For $A_r(S) = 0$ we simply require that $S$ integrate to $0$ over every interval of length $2 r$. If we pick $S$ an integrable function with period $2 r$ then this is quite easy to arrange: almost all such functions integrate to $0$ over such intervals. In particular, $S = \sin(\pi t/r)$ will do.

But now we can add random amounts of $S$ to each $T_\lambda$ thus ensuring that $(T_\lambda)$ does not converge (okay, the amounts are not quite random: we choose them such that the resulting sequence does not converge, but this is always possible). However, since $A_r$ does not "see" $S$, $(A_r(T_\lambda + \alpha_\lambda S))$ still converges.

As I said, I'm not convinced that I understood the question correctly so please do comment if this doesn't look right. But if it doesn't look right, please edit your question so that it's clearer that this isn't what you wanted.