I am trying to prove the following binomial identity:
$$\sum_{i=0}^n (-1)^i\binom{n}{i}\binom{m+i}{m}=(-1)^n\binom{m}{m-n}$$
My idea was to use the identity $$\binom{m}{m-n}=\binom{m}{n}=\sum_{i=0}^n(-1)^i \binom{m+1}{i}$$ but the coefficients of the two sums don't seem to be equal.
I also know there is a combinatorial argument that proves this, but I am trying to find an algebraic proof.
We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k}\tag{1} \end{align*}
We obtain \begin{align*} \color{blue}{\sum_{i=0}^n}&(-1)^i\binom{n}{i}\binom{m+i}{m}\\ &=\sum_{i=0}^{n}(-1)^i\binom{n}{i}[z^m](1+z)^{m+i}\tag{2}\\ &=[z^m](1+z)^m\sum_{i=0}^n\binom{n}{i}(-1)^i(1+z)^i\tag{3}\\ &=[z^m](1+z)^m\left(1-(1+z)\right)^n\tag{4}\\ &=[z^m](1+z)^m(-z)^n\\ &=(-1)^n[z^{m-n}](1+z)^m\tag{5}\\ &\,\,\color{blue}{=(-1)^n\binom{m}{m-n}}\tag{6} \end{align*}
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) we factor out terms independent from $i$.
In (4) we apply the binomial theorem.
In (5) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$>
In (6) we apply (1) again.
Note: This method is referred to as Egorychev I in the publication Egorychev Method: A Hidden Treasure by M. Riedel and H. Mahmoud.
It is unclear what you mean by "algebraic" proof. Most combinatorial identities are found by looking at the coefficients of various polynomials and functions. If you have an exceedingly complex identity it's proof most likely stems from a series of other complicated identities. This one seems fairly straight forward and is a combination of two well-known combinatorial identities.
Let $b_n$ and $a_n$ be sequences of numbers, then $$b_n=\sum_{k=0}^n \binom{n}{k}a_k\ \ \ \ \text{if and only if}\ \ \ \ a_n=\sum_{k=0}^n \binom{n}{k}b_k (-1)^{n-k}$$ This principle follows by multiplying two functions and equating their coefficients $$e^x \sum_{k=0}^\infty \frac{a_k}{k!}x^k = \sum_{k=0}^\infty \frac{b_k}{k!}x^k\ \ \ \ \text{and}\ \ \ \ e^{-x} \sum_{k=0}^\infty \frac{b_k}{k!}x^k = \sum_{k=0}^\infty \frac{a_k}{k!}x^k$$
Vandermonde's identity. Namely, for nonnegative integers $r,n,m$ we have $$\binom{m+n}{r}=\sum_{k=0}^r\binom{m}{k}\binom{n}{r-k}$$ The proof is fairly short involving the coefficients of polynomial products.
Now that we know two of the most basic combinatorial principles (aside from the very basic $\binom{a}{b}=\binom{a}{a-b}$), we can answer your question fairly quickly. In your case we use a Vandermonde's identity with values $m,n,m$ to obtain $$\binom{m+n}{n}=\sum_{i=0}^n\binom{m}{i}\binom{n}{n-i}=\sum_{i=0}^n\binom{n}{i}\binom{m}{m-i}$$ Considering $b_n=\binom{m+n}{n}$ and $a_i=\binom{m}{m-i}$ we can now use principle (1), hence, $$\binom{m}{m-n} = \sum_{i=0}^n \binom{n}{i} \binom{m+i}{i} (-1)^{n-i} = \sum_{i=0}^n \binom{n}{i} \binom{m+i}{m} (-1)^{n-i}$$ In principle you could write these steps out using the functions/polynomials that get multiplied.
Generating function proof.
$$ \begin{split} \sum_{m=0}^{\infty}\sum_{i=0}^n (-1)^i\binom{n}{i}\binom{m+i}{m} x^m &=\sum_{i=0}^n (-1)^i\binom{n}{i}\left(\sum_{m=0}^{\infty}\binom{m+i}{m} x^m \right)\\ &=\sum_{i=0}^n (-1)^i\binom{n}{i}\left(\sum_{m=0}^{\infty}\binom{m+i}{i} x^m \right)\\ &=\sum_{i=0}^n (-1)^i\binom{n}{i}\frac{1}{(1-x)^{i+1}}\\ &=\frac{1}{1-x}\sum_{i=0}^n \binom{n}{i}\left(-\frac{1}{1-x}\right)^i\\ &=\frac{1}{1-x}\left(-\frac{1}{1-x}+1\right)^n\\ &=\frac{(-x)^n}{(1-x)^{n+1}}=(-1)^n\frac{x^n}{(1-x)^{n+1}}\\ &=(-1)^n\sum_{m=0}^{\infty}\binom{m}{n}x^m, \end{split} $$ and therefore, $$ \sum_{i=0}^n (-1)^i\binom{n}{i}\binom{m+i}{m}=(-1)^n\binom{m}{n}. $$
Combinatorial proof (Principle of Inclusion-Exclusion).
Consider the sum $$ \begin{split} (-1)^n\sum_{i=0}^n (-1)^i\binom{n}{i}\binom{m+i}{m} &=\sum_{i=0}^n (-1)^{n-i}\binom{n}{i}\binom{m+i}{m}\\ &=\sum_{i=0}^n (-1)^i\binom{n}{n-i}\binom{m+n-i}{m}\\ &=\sum_{i=0}^n (-1)^i\binom{n}{i}\binom{m+n-i}{m}. \end{split} $$
Consider nondecreasing sequences of $m$ letters from the alphabet $\{1,\dots,n+1\}$. Let $P_i$ ($1\le i\le n$) be the property that the letter $i$ is not in the sequence (so there is no restriction on $n+1$). Let $\Omega=\{P_1,\dots,P_n\}$, and let $S\subseteq\Omega$ be such that $|S|=i$. Then the number $N(\supseteq S)$ of sequences under consideration that have at least all the properties in $S$ is $$ N(\supseteq S)=\binom{m+(n+1-i)-1}{m}=\binom{m+n-i}{m}. $$ For all $i\in\{1,\dots,n+1\}$, let $$ N_i=\sum_{\substack{S\subseteq\Omega\\|S|=i}}N(\supseteq S). $$ Note that $N(\supseteq S)$ only depends on $|S|=i$, so $$ N_i=\binom{n}{i}N(\supseteq S). $$ Then the Inclusion-Exclusion Principle says that the number of such sequences with exactly $0$ properties from $\Omega$ is $$ e_0=\sum_{i=0}^{n}(-1)^iN_i, $$ which is exactly what we want to find.
But the a sequence with exactly $0$ properties in $\Omega$ is simply a nondecreasing sequence on $\{1,\dots,n+1\}$ of length $m$ containing at least $1$ occurrence of each of $1,\dots,n$. Delete the first occurrence of each of them to get an arbitrary a nondecreasing sequence on $\{1,\dots,n+1\}$ of length $m-n$, and the number of such sequences is $$ e_0=\binom{(m-n)+(n+1)-1}{m-n}=\binom{m}{m-n}=\binom{m}{n}. $$ Thus, $$ (-1)^n\sum_{i=0}^n (-1)^i\binom{n}{i}\binom{m+i}{m}=\binom{m}{n}, $$ as desired.
Disclaimer
Not an algebraic proof, at least not fully. But I think it may provide some insights into the solution you are looking for.
Combinatorial Argument
A company is looking to potentially replace some of its employee. At the moment, there are $n$ existing employees and $m$ applicants. The company fires $i$ employees. However, these employees immediately apply to work for the company again, making the number of applicants $m+i$. The company then selects $i$ people out of the $m+i$ applicants to join the company. The number of ways to do this is
$$ \binom{n}{i}\binom{m+i}{i} $$
which is part of the summand on the left-hand side of the question. The question is then to find the difference in the number of possibilities when $i$ is even and $i$ is odd.
Alternative Counting
Let's say that $p$ employees get fired and get hired again, while $q$ employees get replaced by the new applicants. The number of possibilities is
$$ \binom{n-q}{p}\binom{n}{q}\binom{m}{q} $$
with $p+q=i$. Therefore, we have
$$ \sum_{i}^{n}\left(-1\right)^{i}\binom{n}{i}\binom{m+i}{i} = \sum_{p=0}^{n-q}\sum_{q=0}^{n}\left(-1\right)^{p+q}\binom{n-q}{p}\binom{n}{q}\binom{m}{q} $$
remember that for an arbitrary $q\neq n$, we have
$$ \sum_{p=0}^{n-q}\left(-1\right)^{p+q}\binom{n-q}{p}\binom{n}{q}\binom{m}{q}=0 $$
so we have the desired solution
$$ \sum_{i}^{n}\left(-1\right)^{i}\binom{n}{i}\binom{m+i}{i} = \left(-1\right)^{n}\binom{m}{n} $$
