You can calculate/approximate the standard errors via the p-values. First, convert the two-sided p-values into one-sided p-values by dividing them by 2. So you get $p = .0115$ and $p = .007$. Then convert these p-values to the corresponding z-values. For $p = .0115$, this is $z = -2.273$ and for $p = .007$, this is $z = -2.457$ (they are negative, since the odds ratios are below 1). These z-values are actually the test statistics calculated by taking the log of the odds ratios divided by the corresponding standard errors (i.e., $z = log(OR) / SE$). So, it follows that $SE = log(OR) / z$, which yields $SE = 0.071$ for the first and $SE = .038$ for the second study.
Now you have everything to do a meta-analysis. I'll illustrate how you can do the computations with R, using the metafor package:
library(metafor)
yi <- log(c(.85, .91)) ### the log odds ratios
sei <- c(0.071, .038) ### the corresponding standard errors
res <- rma(yi=yi, sei=sei) ### fit a random-effects model to these data
res
Random-Effects Model (k = 2; tau^2 estimator: REML)
tau^2 (estimate of total amount of heterogeneity): 0 (SE = 0.0046)
tau (sqrt of the estimate of total heterogeneity): 0
I^2 (% of total variability due to heterogeneity): 0.00%
H^2 (total variability / within-study variance): 1.00
Test for Heterogeneity:
Q(df = 1) = 0.7174, p-val = 0.3970
Model Results:
estimate se zval pval ci.lb ci.ub
-0.1095 0.0335 -3.2683 0.0011 -0.1752 -0.0438 **
Note that the meta-analysis is done using the log odds ratios. So, $-0.1095$ is the estimated pooled log odds ratio based on these two studies. Let's convert this back to an odds ratio:
predict(res, transf=exp, digits=2)
pred se ci.lb ci.ub cr.lb cr.ub
0.90 NA 0.84 0.96 0.84 0.96
So, the pooled odds ratio is .90 with 95% CI: .84 to .96.
Work on the log scale for as long as you can and then convert to the scale of odds at the last moment. So compute the confidence interval on the log scale and then convert the limits and the estimate to the odds scale as in your step 2. I suppose you could compute a standard error for the odds using the delta method but the way I suggest is simpler I believe.
Best Answer
With an estimate of the log odds ratio $\hat\omega$ & its standard error $\hat\sigma_{\hat\omega}$ you can use the delta method to get an approximation to the standard error of the odds ratio estimate $\newcommand{\e}{\mathrm{e}}\e^\hat\omega$: $$\newcommand{\Var}{\operatorname{Var}} \newcommand{\dif}{\mathrm{d}} \begin{align} \sqrt{\Var \e^{\hat\omega}} & \approx \sqrt{\left(\left.\frac{\dif \e^x}{\dif x}\right|_\hat\omega\right)^2 \Var \hat\omega }\\ & = \e^{\hat\omega} \hat\sigma_{\hat\omega} \end{align}$$
(That's assuming your estimate of the log odds ratio is consistent—i.e. it would tend to the true (population) value as sample size increased.)