(to long for a comment, so I guess it's an answer)
I'm not sure what makes you assert there's a substantive difference between the two cases. When you use Mann-Whitney for testing location-shift alternatives, the assumption is of identical distributions aside from the possible location shift. It's not actually necessary to assume identical distributions. The Mann-Whitney, is, for example, perfectly appropriate for testing scale shift alternatives, or a host of other alternatives, as long as you can compute the distribution of the test statistic under the null. If your rank-based anova is to have a distribution you can compute under $H_0$, you'll need at least some assumptions for the null case there also.
If your assumptions for both are the same (such as both being applied to shift alternatives) and you compute the null distribution on an ANOVA for 2 groups of ranks correctly, your p-values will be identical to the equivalent two-tailed Mann-Whitney, in the same way that $t^2 = F$ for an ordinary 2 group ANOVA compared to a two-tailed two-sample-t (the version with equal-variance).
if I had two groups and both had different, non-normal distributions, but I only wanted to test for a difference in location what test would be preferable? I was under the impression I could use a t-test on ranks, or a Welch t-test on ranks. However, if these tests are the similar to a M_W U test then I guess this is not the case.
It's somewhat of a tricky question, because if they're different shapes 'location difference' doesn't have an obvious meaning in the way it does when they're the same shape.
If you define some measure of location difference (like difference in means or difference in medians or median of pairwise differences or difference in minimum or whatever) then you can do something with it - e.g. try to compute a resampling based distribution, like a bootstrap distribution. It's important to be clear about what you are prepared to assume though.
A Mann-Whitney can be used for more general alternatives than a simple location shift. e.g. For continuous distributions, you can write the null in the form:
$P(X>Y) = \frac{1}{2}$
and the alternative as
$P(X>Y) \neq \frac{1}{2}\quad$ (for a two tailed test)
or
$P(X>Y) < \frac{1}{2}\quad$ (or "$>$", in either case as a one tailed test)
If I recall correctly, Conover's Practical Nonparametric Statistics presents them this way, for example.
One can construct a confidence interval for the median difference using Walsh averages. It does assume that the distribution of the differences is symmetric, but it seems to be OK here. See this document for an explanation of the procedure. It appears that somebody even wrote a Matlab function for this calculation.
EDIT: implementation in Matlab
The psignrank function in R calculates the cumulative distribution of the Sign-rank test statistic. If Matlab does not have that function, then you probably can't easily calculate an exact confidence interval. However you can calculate an approximate interval as described here. I think that is the calculation attempted at the end of the Matlab code.
Best Answer
You can correct for ties in the Mann-Whitney test, have you considered using this version? Some software, e.g., Statistica calculate this automatically.
A tie correction method can be found on Wikipedia: https://en.wikipedia.org/wiki/Mann–Whitney_U_test
Note, that it is still not ideal since your outcome/dependent variable only has three categories.
An alternative could be multinomial logistic regression but your sample size may be too small for this, especially if you originally planned for a t-test in your sample size calculation.