If memory serves, it appears you have forgotten something in your LR statistic.
The likelihood function under the null is
$$L_{H_0} = \theta^{-n_1-n_2}\cdot \exp\left\{-\theta^{-1}\left(\sum x_i+\sum y_i\right)\right\}$$
and the MLE is
$$\hat \theta_0 = \frac {\sum x_i+\sum y_i}{n_1+n_2} = w_1\bar x +w_2 \bar y, \;\; w_1=\frac {n_1}{n_1+n_2},\;w_2=\frac {n_2}{n_1+n_2}$$
So$$ L_{H_0}(\hat \theta_0) = (\hat \theta_0)^{-n_1-n_2}\cdot e^{-n_1-n_2}$$
Under the alternative, the likelihood is
$$L_{H_1} = \theta_1^{-n_1}\cdot \exp\left\{-\theta_1^{-1}\left(\sum x_i\right)\right\}\cdot \theta_2^{-n_2}\cdot \exp\left\{-\theta_2^{-1}\left(\sum y_i\right)\right\}$$
and the MLE's are
$$\hat \theta_1 = \frac {\sum x_i}{n_1} = \bar x, \qquad \hat \theta_2 = \frac {\sum y_i}{n_2} = \bar y$$
So
$$L_{H_1}(\hat \theta_1,\,\hat \theta_2) = (\hat \theta_1)^{-n_1}(\hat \theta_2)^{-n_2}\cdot e^{-n_1-n_2}$$
Consider the ratio
$$\frac {L_{H_1}(\hat \theta_1,\,\hat \theta_2)}{L_{H_0}(\hat \theta_0)} = \frac {(\hat \theta_0)^{n_1+n_2}}{(\hat \theta_1)^{n_1}(\hat \theta_2)^{n_2}}=\left(\frac {\hat \theta_0}{\hat \theta_1}\right)^{n_1} \cdot \left(\frac {\hat \theta_0}{\hat \theta_2}\right)^{n_2}$$
$$= \left(w_1 + w_2 \frac {\bar y}{\bar x}\right)^{n_1} \cdot \left(w_1\frac {\bar x}{\bar y} + w_2 \right)^{n_2}$$
The sample means are independent -so I believe that you can now finish this.
This is one of the cases that an exact test may be obtained and hence there is no reason to appeal to the asymptotic distribution of the LRT. To see this, begin by writing down the definition of an LRT,
$$L = \frac{ \sup_{\lambda \in \omega} f \left( \mathbf{x}, \lambda \right) }{\sup_{\lambda \in \Omega} f \left( \mathbf{x}, \lambda \right)} \tag{1}$$
where $\omega$ is the set of values for the parameter under the null hypothesis and $\Omega$ the respective set under the alternative hypothesis. Note that $\omega$ here is a singleton, since only one value is allowed, namely $\lambda = \frac{1}{2}$. On the other hand the set $\Omega$ is defined as
$$\Omega = \left\{\lambda: \lambda >0 \right\}$$
as the parameter of the exponential distribution is positive, regardless if it is rate or scale. To obtain the LRT we have to maximize over the two sets, as shown in $(1)$. How do we do that? By maximum likelihood of course.
You have already computed the mle for the unrestricted $ \Omega $ set while there is zero freedom for the set $\omega$: $\lambda$ has to be equal to $\frac{1}{2}$. All you have to do then is plug in the estimate and the value in the ratio to obtain
$$L = \frac{ \left( \frac{1}{2} \right)^n \exp\left\{ -\frac{n}{2} \bar{X} \right\} } { \left( \frac{1}{ \bar{X} } \right)^n \exp \left\{ -n \right\} } $$
and we reject the null hypothesis of $\lambda = \frac{1}{2}$ when $L$ assumes a low value, i.e. when
$$L = \frac{ \left( \frac{1}{2} \right)^n \exp\left\{ -\frac{n}{2} \bar{X} \right\} } { \left( \frac{1}{ \bar{X} } \right)^n \exp \left\{ -n \right\} } \leq c $$
Merging constants, this is equivalent to rejecting the null hypothesis when
$$ \left( \frac{\bar{X}}{2} \right)^n \exp\left\{-\frac{\bar{X}}{2} n \right\} \leq k $$
for some constant $k>0$. This is clearly a function of $\frac{\bar{X}}{2}$ and indeed it is easy to show that that the null hypothesis is then rejected for small or large values of $\frac{\bar{X}}{2}$. You can show this by studying the function
$$ g(t) = t^n \exp\left\{ - nt \right\}$$
noting its critical values etc. All that is left for us to do now, is determine the appropriate critical values for a level $\alpha$ test. That is, determine $k_1$ and $k_2$, such that we reject the null hypothesis when
$$\frac{\bar{X}}{2} \leq k_1 \quad \text{or} \quad \frac{\bar{X}}{2} \geq k_2$$
and this is done with probability $\alpha$. This can be accomplished by considering some properties of the gamma distribution, of which the exponential is a special case. Some transformation might be required here, I leave it to you to decide. Remember, though, this must be done under the null hypothesis.
Best Answer
Hint: the likelihood ratio test is not an exact test. The LRT is based upon the asymptotic distribution of the likelihood ratio statistic, i.e. $2 (\log L_1 - \log L_2) \rightarrow \chi^2_{p-q}$. With $L_1$ and $L_2$ arising from likelihoods in MLE parameters for $p$ and $q$ dimensional supports respectively.
An exact test is based upon the actual distribution of the sufficient statistic. What is the sufficient stat for an exponential distribution (how do you show that?) and what distribution does it take (under the null)?