The second expression (RHS) is wrong, the difference of probabilities are signalling a negative number possibility there. What if $x$ is a very large number? Then, you nearly are subtracting $P(T_1+T_2<\infty)$ from $P(T_1<t)$ which will yield a negative number. Apart from this, you have two ways (maybe more) to solve this problem.
Let's start with your way. Your interpretation of $\{N(t)=1\}$ and $\{U_t>x\}$ are correct. Your LHS expression can also be written as:
$$P(T_1+T_2>t+x|T_1+T_2>t, T_1<t) = \frac{P(T_1+T_2>t+x,T_1<t)}{P(T_1+T_2>t,T_1<t)}$$
$T_1$ and $T_2$ are independent exp. RVs with joint density $\lambda^2e^{-\lambda(t_1+t_2)}$, i.e. multiplication of marginals. For the denominator, just draw a 2D plot with axes $t_1,t_2$, draw a line $t_1+t_2=t$; we are going to integrate the joint PDF in $\{T_1<t \cup T_1+T_2>t\}$, i.e. in English, {the region above the line you draw} U {x axis smaller than $t$} U{between $x,y$ axes}, which boils down to the following integral:
$$\int_{0}^{t}{\int_{t-t_1}^{\infty}{\lambda^2e^{-\lambda(t_1+t_2)}dt_2dt_1}}=\lambda te^{-\lambda t}$$
For the numerator, we draw the line $t_1+t_2=t+x$ and take the region above, instead of $t_1+t_2=t$, and that only changes the inner integrals lower bound to $t+x-t_1$, which results in $\lambda te^{-\lambda (t+x)}$. When, you take the ratio, we're left with $e^{-\lambda x}$, which is your answer.
The easy way: You wait for an event to happen for $y=t-t_1$ seconds/mins or whatever. You don't know $t_1$, but that doesn't matter, there exist a $t_1$. And, you wonder if you're going to wait for additional $x$ secs/mins. Since your RV is exponential, which has the memoryless property, i.e. it doesn't depend on how long it has been since its start or $P(X>a+b|X>a)=e^{-\lambda b}$ mathematically, probability of your additional waiting time, $U_t$, only depends on $x$, i.e. $P(U_t>x|N(t)=1)=e^{-\lambda x}$. By the way, $N(t)=1$ carries no important information. It could also be $N(t)=n$, and your answer wouldn't change.
