I want to show that for pairwise differen $i,j,k$ and the matrix $E_{ij}(a) = I_n + ae_{ij}\in \text{SL}_n(K)$ the following relation holds: $$E_{ij}(ab) = E_{ik}(a)E_{kj}(b)E_{ik}(a)^{-1}E_{kj}(b)^{-1}$$ The relation is easy to show using matrix multiplication if we consider elementary matrices without adding the identity matrix. However, I'm stuck on this problem because I don't know how to handle the inverse of the sums (the last two terms).
As this is a problem from an abstract algebra course, not a linear algebra course, I think I may be missing some trickery to show this relation. Is this the case or am I just missing some way to get past the inverse of sums?
As @Dietrich burde says, the inverse is easy to find. However, you can prove this equality using row operations.
Recall that for an matrix $M$, $E_{ij}(a)M$ is the matrix obtained from $M$ by replacing the $i$-th row $L_i$ by $L_i+aL_j$:$L_i\leftarrow L_i+a L_j$
In particular, multipliying by $E_{ij}(a)^{-1}$ on the left is doing this operation backwards: it corresponds to $L_i\leftarrow L_i-aL_j$
Now $E_{ij}(ab)=E_{ij}(ab)I_n$, so it corresponds to $L_i\leftarrow L_i+abL_j$ from the matrix $I_n$.
Your righthand side corresponds to perform four row operations successively from right to left from the matrix $I_n$
After the first operation, then new $k$-th row is $L'_k=L_k-b L_j$.
After the second one, the new $i$-th row is $L'_i=L_i-aL'_k= L_i-aL_k+abL_j$.
After the third one, the new $k$-th row is $L''_k=L'_k+b L_j=L_k-bL_j+bL_j=L_k$
After the fourth one, the new $i$-th row is $L''_i=L'_i+aL''_k=L_i-aL_k+abL_j+aL_k=L_i+abL_j$.
In the process, the rows of index different from $i$ ansd $k$ were untouched. Now, you see from the previous analysis that the $k$-th row is finally untouched as well, so you reall only performed the operation $L_i\leftarrow L_i+abL_j$, as required.