How to tell if a grammar is LALR(1) formally? There is an “informal” definition of $\operatorname{LR}(k)$ (can be recognised by a parser that looks at $k$ symbols ahead) and a “formal” one (as a property of the set of rightmost derivations possible for the grammar). Is there a similar formal criterion of a grammar being $\operatorname{LALR}(k)$ or specifically $\operatorname{LALR}(1)$?
For context, I am interested in looking into grammar transformations that preserve the property of being $\operatorname{LALR}(1)$ (for example, does eliminating right recursion in an $\operatorname{LALR}(1)$ grammar result in a grammar that is still $\operatorname{LALR}(1)$?)
The answer to your first question is: Not that we know of.
A grammar is $\hbox{LALR}(k)$ if and only if its $\hbox{LALR}(k)$ automaton is deterministic. The only way that we know of checking that a grammar is $\hbox{LALR}(k)$ is to build the automaton, or something that essentially amounts to building the automaton.
The good news is that the key complication is empty derivations. So, for example, if a $\hbox{LL}(1)$ grammar has no nonterminals that have only empty derivations (i.e. if every nonterminal has at least one non-empty derivation), it is $\hbox{LALR}(1)$.
See Beatty, On The Relationship Between LL(1) And LR(1) Grammars for further results. You may find something in there that's useful to you.