In most meta-analysis of odds ratios, the standard errors $se_i$ are based on the log odds ratios $log(OR_i)$. So, do you happen to know how your $se_i$ have been estimated (and what metric they reflect? $OR$ or $log(OR)$)? Given that the $se_i$ are based on $log(OR_i)$, then the pooled standard error (under a fixed effect model) can be easily computed. First, let's compute the weights for each effect size: $w_i = \frac{1}{se_i^2}$. Second, the pooled standard error is $se_{FEM} = \sqrt{\frac{1}{\sum w}}$. Furthermore, let $log(OR_{FEM})$ be the common effect (fixed effect model). Then, the ("pooled") 95% confidence interval is $log(OR_{FEM}) \pm 1.96 \cdot se_{FEM}$.
Update
Since BIBB kindly provided the data, I am able to run the 'full' meta-analysis in R.
library(meta)
or <- c(0.75, 0.85)
se <- c(0.0937, 0.1029)
logor <- log(or)
(or.fem <- metagen(logor, se, sm = "OR"))
> (or.fem <- metagen(logor, se, sm = "OR"))
OR 95%-CI %W(fixed) %W(random)
1 0.75 [0.6242; 0.9012] 54.67 54.67
2 0.85 [0.6948; 1.0399] 45.33 45.33
Number of trials combined: 2
OR 95%-CI z p.value
Fixed effect model 0.7938 [0.693; 0.9092] -3.3335 0.0009
Random effects model 0.7938 [0.693; 0.9092] -3.3335 0.0009
Quantifying heterogeneity:
tau^2 < 0.0001; H = 1; I^2 = 0%
Test of heterogeneity:
Q d.f. p.value
0.81 1 0.3685
Method: Inverse variance method
References
See, e.g., Lipsey/Wilson (2001: 114)