Are there algebraic structures with more than one neutral element and/or more than one inverse element?

They do exist but they are algebraic structures with partial operations, i.e. the multiplication $a*b$ is not defined for all $a,b$. Typical examples are "journeys": you can compose a journey from $x$ to $y$ with a journey from $z$ to $w$ if and only if $y=z$. Standard mathematical examples are categories and groupoids. So a groupoid is sometimes thought of as a "group with many identities", and a group is a groupoid with only one identity.

These ideas lead to double categories and groupoids, which have compositions thought of as in different directions. Double groups are just abelian groups, by what is called the Eckmann-Hilton argument, or interchange law, but double groupoids are quite complicated!

Other interesting examples are inverse semigroups. Consider a set $X$ and the set $I(X)$ of all bijections between subsets of $X$. Clearly there is a composition $f \circ g$ of any $f,g \in I(X)$, but the domain of $f \circ g$ may be smaller than expected and even empty. See the wiki entry for more information. In particular the identity $1_A$ on a subset $A$ of $X$ is associated with the domain $A$.

Sept 21, 2016 I'd like to add the point that my definition of "higher dimensional algebra" is that it is the study of algebraic structures with partial operations whose domains are defined by geometric conditions. This allows for a combination of algebra and geometry, which is exploited in our 2011 book Nonabelian Algebraic Topology.

It depends on how you define the 'neutral' element. If you define it as $e*x=x*e=x$ for every $x$ in your set, where $*$ is the operation with respect to which $e$ is 'neutral', then $e$ is going to be unique.

This just came to my mind:

Take a set $X$ with two elements $a$ and $b$. We want to equip this with an interior multiplication that is associative ($X$ is then called a semigroup), and such that $a$ is neutral on the right but not neutral on the left. Then we already know three of four values: $$aa=a,\hspace{20pt}ba=b\hspace{20pt}ab=a\hspace{20pt}bb=?$$ Now, associativity requires that $(ba)b=bb=b(ab)=ba=b$. So $$aa=a,\hspace{20pt}ba=b\hspace{20pt}ab=a\hspace{20pt}bb=b$$ Or in other words, $(x,y)$ gets mapped to the first entry, so the multiplication is projection on the first coordinate. It is then easy to show that this is indeed associative. As we see, both element $a$ and $b$ are neutral on the right.

A structure can have more than one left neutral element ($e$ with $e\circ x=x$ for all $x$) or more than one right neutral element ($e$ with $x\circ e=x$ for all $x$). For example consider the set of functions $f\colon \mathbb N_0\to\mathbb N$ under composition (using $\mathbb N\subset \mathbb N_0$ of course). Then any $f$ with $f(x)=x$ for all $x\ge1$ (but $f(0)$ arbitrary!) is left-neutral. You can do something similar with right-neutral. On the other hand, if an element $e$ is left-neutral and $e'$ is right-neutral, then necessarily $e=e\circ e'=e'$, so if 'neutral' is defined as 'left and right neutral' then uniqueness follows.

A structure with only one neutral can have several left inverses. For example consider the set of functions $f\colon X\to X$ under composition with identity as neutral (obviously). Then a left inverse of $f$ exists if and only if $f$ is injective, a right inverse exists if and only if $f$ is surjective. Unless $f$ is also bijective (which need not be the case as soon as $X$ is infinite), these inverses are not uniquely determined. Admittedly, this is a structure where not all elements have at least one inverse in the first place, but you get the feeling.