Understanding how this reaction is actually performed in the lab is key to answering your question.
Typically in the malonic ester synthesis, a full equivalent (or 2 equivalents if you are carrying out a dialkylation) of base is used, so little, if any, ester remains with the $\ce{\alpha}$-proton still in place.

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Also, keep in mind that both the alkylation and Claisen reaction have an activation energy. This means that it is usually necessary to heat the solution to bring about reaction (see here for a lab example).
The reaction is carried out by first forming the deprotonated malonate ester at room temperature and then dripping it into a refluxing solution of the alkylating agent ($\ce{RX}$ in the above diagram). This procedure achieves the following:
- concentration of non-deprotonated ester is very low
- deprotonated ester is formed at room temperature, so even if any non-deprotonated ester remains there is not enough energy (at room temperature) to make it over the energy barrier and react
- the deprotonated ester is dripped into the solution of refluxing alkylating agent; therefore, now that there is enough thermal energy to bring about reaction, the relative concentration of alkylating agent (the entire solution) is much higher than the relative concentration of deprotonated ester (added drop by drop) making the desired alkylation reaction much more probable than reaction of deprotonated ester with protonated ester (which, for reasons explained above, is in very low concentration in any case).