The statement

If $(X_k)$ is a sequence of independent r.v.'s uniformly bounded in absolute value by some real $c$ such that $\sum_{k}X_k$ converges a.s., then $\sum_{k}EX_k$ converges

follows almost immediately from the Hoffmann–Jørgensen inequality
$$EM^2\le2.4^2c^2+32t_0^2$$
(see e.g. Proposition 6.8),
where $M:=\sup_k|S_k|$, $S_k:=\sum_{j=1}^k X_j$, and $t_0>0$ is such that $P(M>t_0)<1/32$. Indeed, since $S_\infty$ is finite a.s., we have $M<\infty$ a.s. and hence we can assume that $t_0<\infty$, so that $Var\, S_n\le ES_n^2\le EM^2\le2.4^2c^2+32t_0^2<\infty$ for all natural $n$.

Now Kolmogorov's two-series theorem implies that $\sum_k(X_k-EX_k)$ converges a.s. Since $\sum_k X_k$ converges a.s., we conclude that $\sum_k EX_k$ converges, as desired.

**Extended comment:** As the OP noted in a comment, inequality (2) does appear, as Satz III, in Kolmogorov's paper (of 1928, in German). However, apparently there is a mistake in the proof of this inequality in that paper. Namely, with standard modern notations: $E$ in place of $\mathfrak D$ (the expectation symbol in that paper), $A_k$ in place of $\mathfrak f_k$ (a certain event), and $EX1_{A_k}$ in place of $\mathfrak D_{\mathfrak f_k}(X)$, the equality in line 10 on p. 312 of Kolmogorov's paper becomes
$$|t_k-Et_k1_{A_k}|=|s_k-Es_k1_{A_k}|,$$
which was apparently thought of as the consequence of the equality
$$t_k-Et_k1_{A_k}=s_k-Es_k1_{A_k};\tag{$*$}$$
here, according to p. 310 of Kolmogorov's paper, $t_k=s_k-Es_k$. However, actually
$$t_k-Et_k1_{A_k}=s_k-Es_k1_{A_k}-Es_k(1-P(A_k)).$$
So, ($*$) holds only if $Es_k=0$ or $P(A_k)=1$, which latter is not true in general. Thus, the proof seems to only work if $Es_k=0$ for all $k$. This is the first time in my life where I seem to see a mistake made by Kolmogorov!

It appears that this mistake was (tacitly?) corrected in later publications by symmetrization (which obviously provides for the zero-mean condition). If that correction was made indeed tacitly, it appears to be a disservice to people interested in this matter, as the question on this webpage shows.

Anyhow, the Hoffmann–Jørgensen inequality seems to be a more effective and more transparent way to achieve the desired result, as shown above in this answer.

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