You can solve this with a single source shortest path algorithm for weighted directed graphs which allows negative-weight edges, such as the Bellman–Ford algorithm.
The nodes in the graph are sets of items, the source node is the set of items you start with (possibly the empty set), the edges are either buying an item (the edge weight is the purchase price), selling an item (the edge weight is the negative sale price), or trading a set of items (the edge weight is zero).
The Bellman–Ford algorithm will find paths minimising the cost (i.e. maximising the amount of money you have remaining) to each possible set of items you can end up with. Then you can simply take the minimum cost of a path to any set containing the item you wish to acquire.
It's possible that your economy allows arbitrage, i.e. some sequence of trades, purchases and sales by which you can end up with the same items you started with but with more money. (In your example, you can buy item A for 140 and sell it for 150.) In this case the graph contains a negative-weight cycle, so there is no "best" price for any item because you can exploit the arbitrage opportunity for unbounded profit. If there is a negative-weight cycle, the Bellman–Ford algorithm will detect and report it instead of finding shortest paths.
Thank you for your answer, @kaya3. It sounds like framing the problem this way and using such algorithm would work. I'll start experimenting with an implementation soon. Your last paragraph is actually what was the confusing part of the problem to me initially, and it was the reason I opened this question. I just could not figure it out how to allow for the algorithm to make this "turns", where it ends up with the same item but a little profit. I imagine that for really expensive final items, several of this minor trades should take place