Here's another approach that uses a common trick with characteristic functions to avoid having to work out the sums / integrals.
I'll set $\lambda = 1$ without loss of generality, it simplifies notation and can be put back in in obvious ways in what follows. This means all the "$\lambda$"s below are not related to the $\lambda$ in the problem statement, until the very end where I include it again.
First, note that the definition of $Y_t$ involves the sum of $X_i$ corresponding to $U_i \leq t$. This can be thought of as summing "observed" $X_i$, where an $X_i$ is "observed" with a probability $p = t/T$ that is the same across all $i$.. The number of "observed" $X_i$, label it $n$, is therefore distributed Poisson$(pT)$, which is the same as Poisson$(t)$.
The proof of this is straightforward. The number of observed $X_i$, label it $n$, conditional upon $N$ is clearly distributed Binomial$(N, t/T)$. Now, let's look at the characteristic function (ch.f.) of the Binomial distribution:
$\phi_{n|N}(i\theta) = (1-p+p\text{e}^{i\theta})^N$
We will want to integrate out $N$ w.r.t. the Poisson distribution to get the ch.f. of $n$. The simple way to do this is to note that:
$\phi_{n|N}(i\theta) = \exp(N*\log(1-p+p\text{e}^{i\theta}))$
Writing out the integration (summation) gives us:
$\phi_n(i\theta) = \sum_N \exp(N*\log(1-p+p\text{e}^{i\theta})) p(N|\lambda)$
Looking at this, we can see this will have the same form as the ch.f. of a Poisson distribution ($\exp(\lambda(\text{e}^{i\theta}-1))$), just with $\log(1-p+p\text{e}^{i\theta})$ substituted in wherever $i\theta$ appears in the ch.f. Making this substitution gives us:
$\phi_n(i\theta) = \exp(\lambda \text{e}^{\log(1-p+p\text{e}^{i\theta})} - \lambda)$
which quickly reduces to:
$\phi_n(i\theta) = \exp(\lambda(1-p+p\text{e}^{i\theta}) - \lambda)$
which can be rearranged to:
$\phi_n(i\theta) = \exp(p\lambda(\text{e}^{i\theta}-1))$
which is the ch.f. of a Poisson variate with mean $p\lambda$. Substituting $t/T$ for $p$ and $T$ for $\lambda$ gives us the result.
On to step 2. Now we have the ch.f. of the number of elements in the sum $n$. Let's define $\phi_Y(i\theta)$ as the ch.f. of $Y_t$, $\phi_\Sigma(i\theta)$ as the ch.f. of the sum of $n$ $X_i$ and $\phi_X(i\theta)$ as the ch.f. of a single $X_i$. Since the elements are i.i.d., we know that, conditional upon $n$,
$\phi_\Sigma(i\theta) = \phi_X^n(i\theta) = \exp\{n \log \phi_X(i\theta)\}$
We can apply exactly the same approach as above to integrate out $n$:
$\phi_Y(i\theta) = \sum_n \exp\{n \log \phi_X(i\theta)\} p(n | t)$
where we know that $p(n|t)$ is a Poisson distribution. This will be the ch.f. of a Poisson$(t)$ distribution with $\log \phi_X(i\theta)$ substituted for $i\theta$:
$\phi_Y(i\theta) = \exp\{t(\phi_x(i\theta)-1)\}$
Adding the $\lambda$ from the original problem statement gives the answer:
$\phi_Y(i\theta) = \exp\{t\lambda(\phi_x(i\theta)-1)\}$