Here is an ad-hoc proof in the special case where only the operators $d_1$ and $d_2$ appear.
(My apologies for the multiple edits and a bad mistake in the definition of the normal form in the first version.)

I found it easier to work with conjugate partitions. If $\lambda$ is a partition with conjugate partition $\lambda'$ then $\lambda_i' > {\lambda'}_{i+1}$ if and only if $\lambda$ has a part of size $i$. So when $d_i$ is applied to $\lambda$ it reduces (when possible) $\lambda_i'$ by $1$. Hence if we define $e_i$ to be the operator which removes a box from the $i$th part of a partition, then $e_i$ is $d_i$ conjugated by tranposition, and the $e_i$ obey the relations in the question.

Write $1$ for $e_1$ and $2$ for $e_2$ and let $w$, $w'$ be words of length $n$ in the symbols $1$ and $2$. Say that $w$ is *related* to $w'$ if there is a sequence of applications of the identities $121 = 112$ and $122 = 212$ which transforms $w$ to $w'$. For $j \in \lbrace 1,\ldots,n\rbrace$ define

$$c_j(w) = \bigl| \lbrace i : j \le i \le n, w_i = 1 \rbrace \bigr| - \bigl| \lbrace i : j \le i \le n, w_i = 2 \rbrace \bigr| $$

and let $\ell(w)$ be the maximum value of $c_j(w)$ for $j \in \lbrace 1,\ldots,n,n+1\rbrace$, where we take $c_{n+1}(w) = 0$. (Operators compose from right to left, so we must read $w$ from right to left.)

Consider what happens when we apply $w$ to the two row partition $(n+m,n)$. At some point we will have removed $\ell(w)$ more boxes from row $1$ than from row $2$. Moreover, this is the maximum difference in boxes removed. So we get $0$ if and only if $m < \ell(w)$. This shows that if $w$ and $w'$ are equivalent then $\ell(w) = \ell(w')$.

It is obvious that related words are equivalent. So to complete a circle of implications it suffices to prove that if $\ell(w) = \ell(w')$ then $w$ and $w'$ are related. For this, it is sufficient to prove that $w$ is related to a word in the *normal form*

$$u 1\ldots 1 $$

where the final block of $1$ has length $\ell(w)$, and $u$ is of the form $1\ldots 12\ldots 2$ with at least as many $2$s as $1$s.

We work by induction on $\ell(w)$. First suppose that $\ell(w) = 0$. If $w = 1\ldots 12\ldots 2$ we are done. Otherwise let $w = u 212 v$, where $v$ is a (possibly empty) word consisting only of $2$s. Replace $212$ with $122$ to get the related word $u 122 v$, in which the number of final $2$s has increased by one. After finitely many steps we reach the normal form.

Now suppose that $\ell(w) > 0$. Let $j$ be maximal such that $c_j(w) = 1$. So we have
$$w = z 1 v$$
where $v$ has length $n-j$ and $\ell(v) = 0$. Suppose that $j=n$, so $v$ is empty. Since $\ell(z) = \ell(w) - 1$, by induction, $z$ is related to a word in normal form
$$u 1\ldots 1 $$
where the final block has length $\ell(w)-1$. Applying the corresponding sequence of identities to positions $1$, $\ldots$, $n-1$ of $w$ puts $w$ in normal form.

Finally suppose that $j < n$. Then since $\ell(v) = 0$, by the inductive assumption, $v$ is related to $1212 \ldots 12$. Hence $1v$ is related to $11212 \ldots 12$. Repeatedly replacing $112$ with $121$, working from left to right in $1v$, we see that $1v$ is related to $1212 \ldots 121$. So $w$ is related to a word ending with $1$; this reduces to the case $j = n$ already solved.